Exercise 0.1: Implicit differentiation: eleven direct calculations

For each equation, find $\frac{dy}{dx}$ by implicit differentiation, then give its value at the given point, which lies on the curve.

Three moves, always in the same order: differentiate both sides with respect to $x$ , writing a $y'$ every time you differentiate an expression in $y$ ; gather the $y'$ terms on one side; factor out $y'$ and divide. Substitute the coordinates of the point only at the end, or right after the first line if you only want the value.

a) $x^{3}+y^{3}=9$ , at the point $(1, 2)$ .
b) $x^{2}y+xy^{2}=6$ , at the point $(1, 2)$ .
c) $\sqrt{x}+\sqrt{y}=3$ , at the point $(1, 4)$ .
d) $x\cos y+y\cos x=1$ , at the point $(0, 1)$ .
e) $xe^{y}+y=2$ , at the point $(2, 0)$ .
f) $\ln(x+y)=2x-y$ , at the point $\left(\frac{1}{3}, \frac{2}{3}\right)$ .
g) $x=\sin\left(\frac{y}{x}\right)$ , at the point $\left(\frac{1}{2}, \frac{\pi}{12}\right)$ .
h) $x^{2}-3xy+y^{2}=-1$ , at the point $(1, 1)$ .
i) $\sin(xy)=x$ , at the point $\left(\frac{1}{2}, \frac{\pi}{3}\right)$ .
j) $\sqrt{xy}+y^{2}=3$ , at the point $(4, 1)$ .
k) $\frac{x}{y}+\frac{y}{x}=\frac{5}{2}$ , at the point $(1, 2)$ .

Type your answers, the page tells you right or wrong 0/11

a)

y'

at

(1, 2)

b)

y'

at

(1, 2)

c)

y'

at

(1, 4)

d)

y'

at

(0, 1)

e)

y'

at

(2, 0)

f)

y'

at

\left(\frac{1}{3}, \frac{2}{3}\right)

g)

y'

at

\left(\frac{1}{2}, \frac{\pi}{12}\right)

h)

y'

at

(1, 1)

i)

y'

at

\left(\frac{1}{2}, \frac{\pi}{3}\right)

j)

y'

at

(4, 1)

k)

y'

at

(1, 2)

Show the solution

Answers

a) $y'=-\frac{x^{2}}{y^{2}}$ ; $-\frac{1}{4}$
b) $y'=-\frac{y(2x+y)}{x(x+2y)}$ ; $-\frac{8}{5}$
c) $y'=-\frac{\sqrt{y}}{\sqrt{x}}$ ; $-2$
d) $y'=\frac{y\sin x-\cos y}{\cos x-x\sin y}$ ; $-\cos 1\approx-0.540$
e) $y'=-\frac{e^{y}}{xe^{y}+1}$ ; $-\frac{1}{3}$
f) $y'=\frac{2x+2y-1}{x+y+1}$ ; $\frac{1}{2}$
g) $y'=\frac{y}{x}+x\sec\left(\frac{y}{x}\right)$ ; $\frac{\pi}{6}+\frac{1}{\sqrt{3}}\approx 1.101$
h) $y'=\frac{3y-2x}{2y-3x}$ ; $-1$
i) $y'=\frac{\sec(xy)-y}{x}$ ; $\frac{4}{\sqrt{3}}-\frac{2\pi}{3}\approx 0.215$
j) $y'=-\frac{y}{x+4y\sqrt{xy}}$ ; $-\frac{1}{12}$
k) $y'=\frac{y}{x}$ ; $2$

What costs marks here is almost never the derivative of a standard function; it is the $y'$ forgotten behind a $y$ , or the product rule forgotten in a term that mixes $x$ and $y$ . Before writing anything, underline every term that contains BOTH $x$ and $y$ : each of them is a product or a quotient.

a)

$3x^{2}+3y^{2}y'=0$ , so $y'=-\frac{x^{2}}{y^{2}}$ . At the point $(1, 2)$ : $y'=-\frac{1}{4}$ .

b)

Two products: $2xy+x^{2}y'+y^{2}+2xyy'=0$ , hence $y'(x^{2}+2xy)=-(2xy+y^{2})$ and $y'=-\frac{2xy+y^{2}}{x^{2}+2xy}=-\frac{y(2x+y)}{x(x+2y)}$ . At the point $(1, 2)$ : $y'=-\frac{2\times 4}{1\times 5}=-\frac{8}{5}=-1.6$ . Pitfall: $\frac{d}{dx}(x^{2}y)=x^{2}y'$ , forgetting the term $2xy$ .

c)

$\frac{1}{2\sqrt{x}}+\frac{y'}{2\sqrt{y}}=0$ , so $y'=-\frac{\sqrt{y}}{\sqrt{x}}$ . At the point $(1, 4)$ : $y'=-\frac{2}{1}=-2$ .

d)

Two products, each with a trigonometric factor: $\cos y-x\sin y\cdot y'+y'\cos x-y\sin x=0$ . Gather: $y'(\cos x-x\sin y)=y\sin x-\cos y$ , so $y'=\frac{y\sin x-\cos y}{\cos x-x\sin y}$ . At the point $(0, 1)$ : $y'=\frac{0-\cos 1}{1-0}=-\cos 1\approx-0.540$ .

e)

$e^{y}+xe^{y}y'+y'=0$ , so $y'(xe^{y}+1)=-e^{y}$ and $y'=-\frac{e^{y}}{xe^{y}+1}$ . At the point $(2, 0)$ : $y'=-\frac{1}{3}$ . Pitfall: differentiating $e^{y}$ as $e^{y}$ alone, without the $y'$ .

f)

$\frac{1+y'}{x+y}=2-y'$ . Clear the denominator: $1+y'=(2-y')(x+y)$ , then $y'(1+x+y)=2(x+y)-1$ and $y'=\frac{2x+2y-1}{x+y+1}$ . At the point, $x+y=1$ : $y'=\frac{1}{2}$ . The point satisfies the equation, since $\ln 1=0=\frac{2}{3}-\frac{2}{3}$ .

g)

The chain rule, then the quotient rule inside: $1=\cos\left(\frac{y}{x}\right)\cdot\frac{xy'-y}{x^{2}}$ . So $xy'-y=\frac{x^{2}}{\cos(y/x)}$ and $y'=\frac{y}{x}+x\sec\left(\frac{y}{x}\right)$ . At the point, $\frac{y}{x}=\frac{\pi}{6}$ and $\sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}$ : $y'=\frac{\pi}{6}+\frac{1}{\sqrt{3}}\approx 1.101$ .

h)

$2x-3y-3xy'+2yy'=0$ , so $y'(2y-3x)=3y-2x$ and $y'=\frac{3y-2x}{2y-3x}$ . At the point $(1, 1)$ : $y'=\frac{1}{-1}=-1$ . The term $-3xy$ is a product: $-3(y+xy')$ , and both pieces carry the $-3$ .

i)

$\cos(xy)\cdot(y+xy')=1$ , so $y+xy'=\sec(xy)$ and $y'=\frac{\sec(xy)-y}{x}$ . At the point, $xy=\frac{\pi}{6}$ : $y'=\frac{\frac{2}{\sqrt{3}}-\frac{\pi}{3}}{\frac{1}{2}}=\frac{4}{\sqrt{3}}-\frac{2\pi}{3}\approx 0.215$ .

j)

$\frac{y+xy'}{2\sqrt{xy}}+2yy'=0$ . At the point $(4, 1)$ , $\sqrt{xy}=2$ , so $\frac{1+4y'}{4}+2y'=0$ , that is $1+4y'+8y'=0$ and $y'=-\frac{1}{12}\approx-0.083$ . In general, $y'=-\frac{y}{x+4y\sqrt{xy}}$ . Here substituting early is safer than dragging the square root to the very end.

k)

$\frac{y-xy'}{y^{2}}+\frac{xy'-y}{x^{2}}=0$ , that is $(xy'-y)\left(\frac{1}{x^{2}}-\frac{1}{y^{2}}\right)=0$ . At the point $(1, 2)$ , the second bracket equals $\frac{3}{4}\neq 0$ , so $xy'-y=0$ and $y'=\frac{y}{x}=2$ . Explanation: multiplying the equation by $2xy$ gives $(2x-y)(x-2y)=0$ , that is, the two lines $y=2x$ and $y=\frac{x}{2}$ , and the slope of a line is constant.

Check: on every line, the given point satisfies the original equation, and the slope found agrees with $y'=-\frac{F_{x}}{F_{y}}$ computed on a calculator, where $F_{x}$ and $F_{y}$ are the rates of change of the left side minus the right side with respect to $x$ alone and to $y$ alone.

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Exercise 1: The principle: differentiate an equation, not a function

Until now $y$ was given by a formula. Here it is only CONSTRAINED by an equation, and yet its derivative can be computed. The thread of the whole set fits in one sentence: $y$ is a function of $x$ that we do not know, so every $y$ we differentiate leaves a $y'$ behind.

a) Differentiate both sides of $x^{2}+y^{2}=25$ with respect to $x$ and solve for $y'$ .
b) Give the equation of the tangent line at the point (3, 4), and check on the figure that it is perpendicular to the radius.
c) Redo the computation by the EXPLICIT method, setting $y=\sqrt{25-x^{2}}$ , and compare the two results at $x=3$ .
d) What does the implicit formula give at the point (3, -4)? Why is the explicit method less convenient here?
e) A student writes “ $2x+2y=0$ , so $y=-x$ ”. Where is the error, and why is the conclusion absurd on this circle?

Show the solution

Answers

a) $y'=-\frac{x}{y}$
b) $y=-\frac{3}{4}x+\frac{25}{4}$ , perpendicular to the radius
c) $-\frac{3}{4}$ by both methods
d) $y'=\frac{3}{4}$ at $(3, -4)$
e) $y^{2}$ differentiated without the factor $y'$

a)

Differentiate term by term, remembering that $y$ depends on $x$ : $\frac{d}{dx}(x^{2})=2x$ , and $\frac{d}{dx}(y^{2})=2y\cdot y'$ by the chain rule, since $y^{2}$ is a square whose inside is a function of $x$ . The right side is a constant, with derivative zero. Hence $2x+2yy'=0$ , then $y'=-\frac{x}{y}$ for $y\neq 0$ . Note the form of the result: it depends on $x$ AND $y$ , which is normal, since the same $x$ can correspond to two points of the circle.

b)

At (3, 4): $y'=-\frac{3}{4}$ . The tangent line is $y=4-\frac{3}{4}(x-3)$ , that is $3x+4y=25$ . The radius joining the origin to (3, 4) has slope $\frac{4}{3}$ , and $\left(-\frac{3}{4}\right)\times\frac{4}{3}=-1$ : the two lines are indeed perpendicular, as the figure shows. This is not a coincidence of this point; it holds everywhere on the circle, since $y'=-\frac{x}{y}$ is always the negative reciprocal of $\frac{y}{x}$ .

c)

On the upper half, $y=(25-x^{2})^{1/2}$ , so $y'=\frac{1}{2}(25-x^{2})^{-1/2}\cdot(-2x)=\frac{-x}{\sqrt{25-x^{2}}}$ . At $x=3$ : $\frac{-3}{\sqrt{16}}=-\frac{3}{4}$ , the same as the implicit result. The two methods say the same thing, and we can see why: $\sqrt{25-x^{2}}$ is nothing but $y$ , so the explicit formula IS the implicit formula with $y$ substituted.

d)

The implicit formula directly gives $y'=-\frac{3}{-4}=\frac{3}{4}$ : the tangent rises, which is correct on the lower half of the circle. The explicit method first requires SWITCHING formulas, $y=-\sqrt{25-x^{2}}$ , because the square root only describes the upper half. This is the decisive advantage of the implicit method: a single formula covers the whole curve, including the portions that are not the graph of any function.

e)

The error is differentiating $y^{2}$ as if $y$ were a constant, or differentiating it “with respect to $y$ ”: this gives $2y$ instead of $2yy'$ . The conclusion $y=-x$ is a LINE, whereas we were looking for a slope; the question has been replaced by another one. Immediate test: this line claims to pass through (3, -3), which is not on the circle since $9+9=18\neq 25$ . A result of implicit differentiation that no longer contains $y'$ is always wrong.

Exercise 2: The five structures that leave a y prime

The whole technique comes down to recognizing five forms. As soon as a $y$ appears inside an operation, the chain rule applies and a $y'$ comes out; as soon as $x$ and $y$ meet in the same term, the product or quotient rule takes over.

a) Differentiate with respect to $x$ : $y^{3}$ , $\sqrt{y}$ , $\frac{1}{y}$ .
b) Differentiate with respect to $x$ : $\sin y$ , $e^{y}$ , $\ln y$ .
c) Differentiate with respect to $x$ : $xy$ , $x^{2}y^{3}$ , $\frac{x}{y}$ .
d) Differentiate the equation $x^{2}y+\sin(xy)=4$ and solve for $y'$ .
e) In $e^{xy}=x+y$ , solve for $y'$ , then give its value at the point (0, 1).
f) Differentiate $x\sin y+y\cos x=\frac{\pi}{2}$ , check that the point $\left(0, \frac{\pi}{2}\right)$ is on the curve and give $y'$ at that point. Then differentiate $\tan(x+y)=x$ and express $y'$ in terms of $x$ ALONE.
g) In $x^{2}y=\cos(x^{2}y)$ , differentiate both sides, factor and deduce $y'$ . Then explain the result by setting $u=x^{2}y$ : what is the value of $u$ ?

Type your answers, the page tells you right or wrong 0/15

a)

\frac{d}{dx}(y^{3})

\frac{d}{dx}(\sqrt{y})

\frac{d}{dx}\left(\frac{1}{y}\right)

b)

\frac{d}{dx}(\sin y)

\frac{d}{dx}(e^{y})

\frac{d}{dx}(\ln y)

c)

\frac{d}{dx}(xy)

\frac{d}{dx}(x^{2}y^{3})

\frac{d}{dx}\left(\frac{x}{y}\right)

d)

y'

e)

y'

at

(0, 1)

f)

y'

at

\left(0, \frac{\pi}{2}\right)

y'

for

\tan(x+y)=x

g)

y'

u=x^{2}y

Show the solution

Answers

a) $3y^{2}y'$ ; $\frac{y'}{2\sqrt{y}}$ ; $-\frac{y'}{y^{2}}$
b) $y'\cos y$ ; $e^{y}y'$ ; $\frac{y'}{y}$
c) $y+xy'$ ; $2xy^{3}+3x^{2}y^{2}y'$ ; $\frac{y-xy'}{y^{2}}$
d) $y'=\frac{-y\left(2x+\cos(xy)\right)}{x\left(x+\cos(xy)\right)}$
e) $y'=\frac{1-ye^{xy}}{xe^{xy}-1}$ ; $y'=0$ at $(0, 1)$
f) $y'=\frac{y\sin x-\sin y}{x\cos y+\cos x}$ , $y'=-1$ at $\left(0, \frac{\pi}{2}\right)$ ; $y'=-\frac{x^{2}}{1+x^{2}}$
g) $y'=-\frac{2y}{x}$ ; $u=x^{2}y\approx 0.739$ , constant

a)

$\frac{d}{dx}(y^{3})=3y^{2}y'$ ; $\frac{d}{dx}(\sqrt{y})=\frac{y'}{2\sqrt{y}}$ ; $\frac{d}{dx}\left(\frac{1}{y}\right)=-\frac{y'}{y^{2}}$ . In all three cases, we differentiated the outer function keeping $y$ inside, then multiplied by $y'$ , the derivative of the inside. This is the chain rule, with no adjustment whatsoever.

b)

$\frac{d}{dx}(\sin y)=\cos y\cdot y'$ ; $\frac{d}{dx}(e^{y})=e^{y}y'$ ; $\frac{d}{dx}(\ln y)=\frac{y'}{y}$ . Notice the last one: it is exactly logarithmic differentiation, where $\frac{y'}{y}$ appears naturally. An equation containing $\ln y$ is therefore differentiated with no extra effort.

c)

$\frac{d}{dx}(xy)=1\cdot y+x\cdot y'=y+xy'$ : this is the PRODUCT rule, and forgetting the term $y$ is the second most frequent mistake of the chapter. $\frac{d}{dx}(x^{2}y^{3})=2xy^{3}+x^{2}\cdot 3y^{2}y'=2xy^{3}+3x^{2}y^{2}y'$ , product rule then chain rule. $\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{1\cdot y-x\cdot y'}{y^{2}}=\frac{y-xy'}{y^{2}}$ , quotient rule.

d)

Term by term: $2xy+x^{2}y'+\cos(xy)\cdot(y+xy')=0$ . Gather everything that carries $y'$ on one side: $y'\left(x^{2}+x\cos(xy)\right)=-2xy-y\cos(xy)$ , so $y'=\frac{-2xy-y\cos(xy)}{x^{2}+x\cos(xy)}=\frac{-y\left(2x+\cos(xy)\right)}{x\left(x+\cos(xy)\right)}$ . The method is always the same three moves: differentiate, gather the $y'$ terms, factor. Never try to solve for $y$ first; it would often be impossible.

e)

$e^{xy}\cdot(y+xy')=1+y'$ , that is $ye^{xy}+xy'e^{xy}=1+y'$ , then $y'\left(xe^{xy}-1\right)=1-ye^{xy}$ and $y'=\frac{1-ye^{xy}}{xe^{xy}-1}$ . The point (0, 1) is indeed on the curve: $e^{0}=1$ and $0+1=1$ ✓. There we find $y'=\frac{1-1\cdot 1}{0-1}=\frac{0}{-1}=0$ : a horizontal tangent. Numerical check: for $x=0.1$ , the equation $e^{0.1y}=0.1+y$ gives $y\approx 1.0058$ , a change of $0.006$ for a step of $0.1$ : the curve is almost flat at this point, as $y'=0$ predicts.

f)

Two products, each with a hidden $y'$ : $\sin y+x\cos y\cdot y'+y'\cos x-y\sin x=0$ . Gather: $y'\left(x\cos y+\cos x\right)=y\sin x-\sin y$ , so $y'=\frac{y\sin x-\sin y}{x\cos y+\cos x}$ . The point is on the curve: $0\cdot\sin\frac{\pi}{2}+\frac{\pi}{2}\cos 0=\frac{\pi}{2}$ ✓, and $y'=\frac{0-1}{0+1}=-1$ . For the second equation, the chain rule acts on $x+y$ , whose derivative is $1+y'$ and not $y'$ : $\sec^{2}(x+y)\cdot(1+y')=1$ , hence $y'=\cos^{2}(x+y)-1=-\sin^{2}(x+y)$ . The identity $\sec^{2}=1+\tan^{2}$ and the equation itself, $\tan(x+y)=x$ , give $\sec^{2}(x+y)=1+x^{2}$ , so $y'=\frac{1}{1+x^{2}}-1=-\frac{x^{2}}{1+x^{2}}$ . The $y$ has disappeared: the original equation is also useful AFTER differentiating. We recognize the derivative of $\arctan x-x$ , and that is indeed what $y$ equals on the branch through the origin.

g)

The product $x^{2}y$ appears on both sides, so its derivative $2xy+x^{2}y'$ does too: $2xy+x^{2}y'=-\sin(x^{2}y)\cdot\left(2xy+x^{2}y'\right)$ , that is $\left(2xy+x^{2}y'\right)\left(1+\sin(x^{2}y)\right)=0$ . The second factor never vanishes on the curve (see below), so $2xy+x^{2}y'=0$ and $y'=-\frac{2y}{x}$ , with no cosine or sine. The trick explains it: with $u=x^{2}y$ , the equation becomes $u=\cos u$ . The function $u-\cos u$ has derivative $1+\sin u\ge 0$ , so it is increasing and has only one zero, at $u\approx 0.739$ . The product $x^{2}y$ is CONSTANT, the curve is $y=\frac{0.739}{x^{2}}$ , and $\sin(0.739)\approx 0.674\neq -1$ confirms that the factor $1+\sin u$ does not vanish. An equation in which $x$ and $y$ only appear through one combination fixes that combination: you can see it before differentiating, and save the whole computation.

Exercise 3: The folium of Descartes, a curve that is the graph of no function

The curve with equation $x^{3}+y^{3}=6xy$ passes three times above some $x$ -values. No formula $y=f(x)$ describes it, and yet its tangent can be computed everywhere.

a) Check that the point (3, 3) is on the curve, then compute $y'$ in terms of $x$ and $y$ .
b) Give the slope of the tangent at (3, 3) and the equation of this tangent line.
c) Find the point of the loop where the tangent is HORIZONTAL.
d) Deduce from it, without any new computation, the point where the tangent is VERTICAL. What argument do you use?
e) Why does the vertical line test fail on this curve, and why does that not get in the way of implicit differentiation?

Show the solution

Answers

a) $y'=\frac{2y-x^{2}}{y^{2}-2x}$
b) Slope $-1$ ; $x+y=6$
c) $\left(\sqrt[3]{16}, \sqrt[3]{32}\right)\approx(2.520, 3.175)$
d) $(3.175, 2.520)$ by symmetry
e) Three points on $x=2$ : not a function

a)

$3^{3}+3^{3}=54$ and $6\times 3\times 3=54$ ✓. Differentiate: $3x^{2}+3y^{2}y'=6y+6xy'$ . Gather: $y'\left(3y^{2}-6x\right)=6y-3x^{2}$ , so $y'=\frac{6y-3x^{2}}{3y^{2}-6x}=\frac{2y-x^{2}}{y^{2}-2x}$ after dividing by 3.

b)

At (3, 3): $y'=\frac{6-9}{9-6}=\frac{-3}{3}=-1$ . The tangent line is $y=3-1\cdot(x-3)$ , that is $x+y=6$ . The figure shows it: it touches the loop at its point farthest from the origin, which is consistent with the symmetry of the equation under swapping $x$ and $y$ , of which (3, 3) is the fixed point.

c)

The tangent is horizontal when the numerator is zero while the denominator is not: $2y-x^{2}=0$ , that is $y=\frac{x^{2}}{2}$ . Substitute into the equation of the curve: $x^{3}+\frac{x^{6}}{8}=6x\cdot\frac{x^{2}}{2}=3x^{3}$ , hence $\frac{x^{6}}{8}=2x^{3}$ and, for $x\neq 0$ , $x^{3}=16$ . So $x=16^{1/3}\approx 2.5198$ and $y=\frac{x^{2}}{2}\approx 3.1748$ . The denominator there equals $y^{2}-2x\approx 10.079-5.040=5.039\neq 0$ , so the conclusion is valid.

d)

The equation is SYMMETRIC under swapping $x$ and $y$ : $x^{3}+y^{3}=6xy$ remains true if the two letters are exchanged. The curve is therefore invariant under reflection in the line $y=x$ , and this reflection exchanges horizontal and vertical. The point with a vertical tangent is thus the image of the previous one: $x\approx 3.1748$ and $y\approx 2.5198$ . We check it on the formula: the denominator $y^{2}-2x\approx 6.349-6.350\approx 0$ , while the numerator does not vanish. Using a symmetry saves half the work here.

e)

The vertical line test requires each $x$ -value to give at most one $y$ -value; here the vertical line $x=2$ meets the curve at three points, so the curve is the graph of no function. That prevents nothing: implicit differentiation does not assume that the WHOLE curve is a graph, only that, AROUND THE POINT considered, it can be seen locally as $y=f(x)$ . This is exactly what the condition “denominator not zero” guarantees, and it is why the method breaks down precisely at points with a vertical tangent.

Exercise 4: Horizontal and vertical tangents: reading the two zeros

An implicit derivative almost always comes as a quotient. The numerator controls the horizontal tangents, the denominator the vertical tangents, and the case where both vanish together signals a singular point.

a) For the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ , compute $y'$ and give all the points with a horizontal tangent, then with a vertical tangent.
b) Same work for the tilted ellipse $x^{2}+xy+y^{2}=7$ .
c) For the curve $y^{2}=x^{3}$ , compute $y'$ and study what happens at the origin.
d) State the general rule in one sentence, and say what must be checked before concluding.

Show the solution

Answers

a) Horizontal at $(0, \pm 2)$ ; vertical at $(\pm 3, 0)$
b) Horizontal at $\pm(1.528, -3.055)$ ; vertical at $\pm(-3.055, 1.528)$
c) $y'=\frac{3x^{2}}{2y}$ : $\frac{0}{0}$ , cusp
d) Check the point, and that the other term is not zero

a)

$\frac{2x}{9}+\frac{2yy'}{4}=0$ gives $y'=-\frac{4x}{9y}$ . Horizontal tangent when $x=0$ and $y\neq 0$ : the points (0, 2) and (0, -2), the endpoints of the minor axis. Vertical tangent when $y=0$ : the points (3, 0) and (-3, 0), the endpoints of the major axis. This matches intuition: an ellipse is flat at the top and bottom, vertical on the left and right.

b)

$2x+\left(y+xy'\right)+2yy'=0$ , so $y'\left(x+2y\right)=-(2x+y)$ and $y'=-\frac{2x+y}{x+2y}$ . Horizontal: $2x+y=0$ , that is $y=-2x$ ; substituting, $x^{2}-2x^{2}+4x^{2}=7$ gives $3x^{2}=7$ and $x=\pm\sqrt{\frac{7}{3}}\approx\pm 1.5275$ , hence the point $(1.5275, -3.0551)$ and its opposite. Vertical: $x+2y=0$ , that is $x=-2y$ , and by the symmetry of the equation in $x$ and $y$ we get the symmetric point $(-3.0551, 1.5275)$ and its opposite. The ellipse is tilted, so its horizontal tangents are not on the axes.

c)

$2yy'=3x^{2}$ gives $y'=\frac{3x^{2}}{2y}$ . At the origin, numerator AND denominator vanish: the formula decides nothing. Switching to the explicit form, $y=\pm x^{3/2}$ for $x\geq 0$ , so $y'=\pm\frac{3}{2}\sqrt{x}\to 0$ as $x\to 0^{+}$ : both branches reach the origin with zero slope, but they form a SPIKE, a cusp. The curve is not smooth there; it is a singular point, and the only way to know it is to have noticed the double zero.

d)

Rule: the tangent is horizontal where the numerator of $y'$ is zero while the denominator is not, and vertical where the denominator is zero while the numerator is not. Before concluding, two compulsory checks: that the point really belongs to the curve, and that the other term of the quotient is not zero as well. The second one is the one students skip, and it is what distinguishes a genuine tangent from a cusp.

Exercise 5: The implicit second derivative and concavity

The second differentiation is done exactly like the first, with one discipline: the $y'$ that appeared in the first round is itself a function of $x$ , so it is differentiated in turn. Then $y'$ is replaced by its expression, never before.

a) On the circle $x^{2}+y^{2}=25$ , we found $y'=-\frac{x}{y}$ . Compute $y''$ and simplify it using the equation of the circle.
b) Give $y''$ at the point (3, 4) and at the point (3, -4). Interpret the signs.
c) For $x^{2}+xy=6$ , compute $y'$ , then $y''$ .
d) A student differentiates $y'=-\frac{x}{y}$ and writes $y''=-\frac{1}{y}$ . Which rule has been forgotten, and how far off is the answer at (3, 4)?

Show the solution

Answers

a) $y''=-\frac{25}{y^{3}}$
b) $-\frac{25}{64}$ and $\frac{25}{64}$
c) $y'=-\frac{2x+y}{x}$ , $y''=\frac{2x+2y}{x^{2}}$
d) Product rule forgotten: $-0.25$ against $-0.391$

a)

Differentiate $y'=-xy^{-1}$ as a product: $y''=-\left(1\cdot y^{-1}+x\cdot(-y^{-2})y'\right)=-\frac{1}{y}+\frac{xy'}{y^{2}}$ . NOW replace $y'$ by $-\frac{x}{y}$ : $y''=-\frac{1}{y}-\frac{x^{2}}{y^{3}}=-\frac{y^{2}+x^{2}}{y^{3}}$ . Finally use the original equation, $x^{2}+y^{2}=25$ , which gives the compact form $y''=-\frac{25}{y^{3}}$ . Substituting the equation of the curve at the end is what turns an unreadable expression into a usable result.

b)

At (3, 4): $y''=-\frac{25}{64}\approx-0.391$ , negative, so the curve is CONCAVE DOWN: that is indeed the shape of the upper half of the circle. At (3, -4): $y''=-\frac{25}{-64}=+\frac{25}{64}\approx 0.391$ , positive, the lower half is concave up. The formula $-\frac{25}{y^{3}}$ says it at a glance: the sign of $y''$ is the opposite of the sign of $y$ , so the circle changes concavity exactly when going from one half to the other.

c)

$2x+y+xy'=0$ gives $y'=-\frac{2x+y}{x}$ . Differentiate this quotient: $y''=-\frac{\left(2+y'\right)x-\left(2x+y\right)\cdot 1}{x^{2}}=-\frac{2x+xy'-2x-y}{x^{2}}=-\frac{xy'-y}{x^{2}}$ . Substitute $xy'=-(2x+y)$ : $y''=-\frac{-(2x+y)-y}{x^{2}}=\frac{2x+2y}{x^{2}}$ . Check at the point (1, 5), which is on the curve since $1+5=6$ : $y'=-7$ and $y''=\frac{2+10}{1}=12$ .

d)

The PRODUCT rule has been forgotten: $-\frac{x}{y}$ is the product of $-x$ and $\frac{1}{y}$ , and $\frac{1}{y}$ depends on $x$ . Only the factor $-x$ was differentiated. At (3, 4), the student's answer is $-\frac{1}{4}=-0.25$ against the true value $-\frac{25}{64}\approx-0.391$ : an error of 36 percent, which does not change the sign and therefore goes unnoticed in a concavity test, but ruins any computation of a radius of curvature or of a second-order approximation.

Exercise 6: Two families of curves that always cross at right angles

Two families are called orthogonal when every curve of one crosses every curve of the other perpendicularly. This is the structure of field lines and equipotential lines in electricity, and implicit differentiation proves it in three lines.

a) Compute the slope at any point of a curve of the family $xy=k$ , then of a curve of the family $x^{2}-y^{2}=c$ .
b) Prove that the two families are orthogonal at every point of intersection.
c) Check it on the intersection of $xy=2$ and $x^{2}-y^{2}=3$ : find the point in the first quadrant, then the two slopes.
d) Why is the proof in part b stronger than a check like the one in part c?
e) The two families share a problematic point. Which one, and what happens there?

Show the solution

Answers

a) $-\frac{y}{x}$ and $\frac{x}{y}$
b) Product of the slopes $-1$
c) $(2, 1)$ : slopes $-\frac{1}{2}$ and $2$
d) Generic point: every intersection
e) The origin: an angle of $45°$

a)

On $xy=k$ : $y+xy'=0$ , so $y'=-\frac{y}{x}$ . On $x^{2}-y^{2}=c$ : $2x-2yy'=0$ , so $y'=\frac{x}{y}$ . Note that $k$ and $c$ have disappeared: the slope depends only on the POINT, not on the particular curve of the family that passes through it. This is exactly what makes a general proof possible.

b)

At an intersection point $(x_{0}, y_{0})$ with $x_{0}\neq 0$ and $y_{0}\neq 0$ , the product of the two slopes is $\left(-\frac{y_{0}}{x_{0}}\right)\times\frac{x_{0}}{y_{0}}=-1$ . Two lines whose slopes have product $-1$ are perpendicular, so the two curves cross at right angles. The conclusion depends neither on $k$ , nor on $c$ , nor on the point: it holds for every intersection of every pair of curves from the two families.

c)

From $xy=2$ we get $y=\frac{2}{x}$ , which we substitute: $x^{2}-\frac{4}{x^{2}}=3$ , that is $x^{4}-3x^{2}-4=0$ . Setting $X=x^{2}$ : $X^{2}-3X-4=0$ , with roots $X=4$ and $X=-1$ , of which only the first is acceptable. So $x=2$ and $y=1$ . The slopes are $-\frac{1}{2}$ for the hyperbola $xy=2$ and $\frac{2}{1}=2$ for $x^{2}-y^{2}=3$ , with product $-1$ ✓. On the figure, it is the point where the lower solid curve meets the lower dashed curve.

d)

The check in part c proves nothing beyond the point (2, 1): it is compatible with the idea that orthogonality might be a coincidence of that particular point. The proof in part b, on the other hand, deals with a GENERIC point and establishes the result for infinitely many intersections at once. This is the specific benefit of implicit differentiation in this problem: because it gives the slope in terms of the coordinates and not of the curve, it lets us reason about a whole family at once.

e)

The origin. It belongs to the family $xy=k$ for $k=0$ (the union of the two axes) and to the family $x^{2}-y^{2}=c$ for $c=0$ (the union of the two diagonal lines through the origin). Both slope formulas are indeterminate there, of the form $\frac{0}{0}$ , and the degenerate curves through it do not cross at right angles: the axes and the diagonals make an angle of 45 degrees. Orthogonality therefore holds everywhere EXCEPT where the reasoning was not allowed, which is the usual situation and not an awkward exception.

Exercise 7: Conics and the point form of the tangent

There is a substitution rule that gives the tangent to a conic without differentiating: replace $x^{2}$ by $xx_{0}$ and $y^{2}$ by $yy_{0}$ . It looks like magic; implicit differentiation proves it.

a) For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , compute $y'$ at the point $(x_{0}, y_{0})$ of the curve, then prove that the tangent line can be written $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=1$ .
b) Apply it to $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ at the point with $x$ -coordinate 4 and a positive $y$ -coordinate: find the point, the slope and the tangent line.
c) State the analogous formula for the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and check it on $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ at the point (5, 2.25).
d) Does the point form still work at points with a vertical tangent, where $y'$ does not exist? Justify.

Show the solution

Answers

a) $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=1$
b) $(4, 1.8)$ , slope $-0.8$ , $4x+5y=25$
c) $\frac{xx_{0}}{a^{2}}-\frac{yy_{0}}{b^{2}}=1$ ; slope $1.25$
d) Yes: $x=a$ at $(a, 0)$

a)

$\frac{2x}{a^{2}}+\frac{2yy'}{b^{2}}=0$ gives $y'=-\frac{b^{2}x}{a^{2}y}$ , so at the point of contact $y'=-\frac{b^{2}x_{0}}{a^{2}y_{0}}$ for $y_{0}\neq 0$ . The tangent line is $y-y_{0}=-\frac{b^{2}x_{0}}{a^{2}y_{0}}(x-x_{0})$ . Multiply by $\frac{y_{0}}{b^{2}}$ : $\frac{yy_{0}-y_{0}^{2}}{b^{2}}=-\frac{x_{0}x-x_{0}^{2}}{a^{2}}$ , then regroup: $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}$ . Now the right side equals 1, BECAUSE the point of contact is on the ellipse. Hence the announced form. This last step explains the substitution rule: it encodes the fact that the point of contact belongs to the curve.

b)

$\frac{16}{25}+\frac{y^{2}}{9}=1$ gives $\frac{y^{2}}{9}=\frac{9}{25}$ , so $y^{2}=\frac{81}{25}$ and $y_{0}=\frac{9}{5}=1.8$ . The slope is $-\frac{9\times 4}{25\times 1.8}=-\frac{36}{45}=-0.8$ . Tangent line by the point form: $\frac{4x}{25}+\frac{1.8y}{9}=1$ , that is $\frac{4x}{25}+\frac{y}{5}=1$ , or $4x+5y=25$ . Check: the slope of $4x+5y=25$ is $-\frac{4}{5}=-0.8$ ✓, and the point (4, 1.8) satisfies it since $16+9=25$ ✓.

c)

For the hyperbola, the only change is the sign: $\frac{xx_{0}}{a^{2}}-\frac{yy_{0}}{b^{2}}=1$ , and differentiation gives $y'=\frac{b^{2}x}{a^{2}y}$ , without the minus sign. The point (5, 2.25) is indeed on the curve: $\frac{25}{16}-\frac{5.0625}{9}=1.5625-0.5625=1$ ✓. Slope: $\frac{9\times 5}{16\times 2.25}=\frac{45}{36}=1.25$ . Point form of the tangent: $\frac{5x}{16}-\frac{2.25y}{9}=1$ , that is $\frac{5x}{16}-\frac{y}{4}=1$ , whose slope is $\frac{5}{16}\times 4=1.25$ ✓.

d)

Yes, and that is its main advantage. At the point $(a, 0)$ of the ellipse, $y'$ does not exist because the denominator $a^{2}y_{0}$ is zero; but the point form gives $\frac{xa}{a^{2}}+0=1$ , that is $x=a$ , which is exactly the expected vertical tangent. The reason is that the point form never divides by $y_{0}$ in its final statement: the division only appears in the middle of the proof, and the result itself remains valid by continuity. A formula can therefore be more general than the method that produced it, which is worth noticing.

Exercise 8: Five statements to correct

Each of the following five statements is FALSE. Say why, and give the correct statement.

1) “Differentiating $x^{2}+y^{2}=25$ with respect to $x$ gives $2x+2y=0$ .”
2) “ $\frac{d}{dx}(xy)=y'$ , since the derivative of $x$ is 1.”
3) “A properly computed implicit derivative must depend on $x$ only: if $y$ is still there, the work is not finished.”
4) “Since $y'=-\frac{x}{y}$ is a fraction defined as soon as $y\neq 0$ , the curve never has a vertical tangent.”
5) “To differentiate $y^{3}$ with respect to $x$ , write $3y^{2}$ , just as you would write $3x^{2}$ for $x^{3}$ .”

Show the solution

Answers

1) $2x+2yy'=0$
2) $\frac{d}{dx}(xy)=y+xy'$
3) $y'$ normally depends on $x$ and $y$
4) Vertical tangents at $(\pm 5, 0)$
5) $\frac{d}{dx}(y^{3})=3y^{2}y'$

1)

FALSE. The term $y^{2}$ is a composition: $y$ depends on $x$ , so the chain rule requires $\frac{d}{dx}(y^{2})=2yy'$ . The correct differentiation is $2x+2yy'=0$ , hence $y'=-\frac{x}{y}$ . The test that never fails: an implicit differentiation whose result contains NO $y'$ has not differentiated with respect to the right variable. Here the statement leads to $y=-x$ , which describes a line meeting the circle at only two points.

2)

FALSE. $xy$ is a PRODUCT of two functions of $x$ , so $\frac{d}{dx}(xy)=1\cdot y+x\cdot y'=y+xy'$ . The forgotten term $y$ is the one that comes from the derivative of the factor $x$ . Numerical consequence on $xy=6$ at the point (2, 3): the correct derivative gives $3+2y'=0$ , so $y'=-1.5$ , whereas the statement would give $y'=0$ , a horizontal tangent on a hyperbola that has none.

3)

FALSE. It is the opposite: an implicit derivative NORMALLY depends on $x$ and $y$ , and that is precisely what lets it describe a curve that passes several times above the same $x$ -value. On the circle, $y'=-\frac{x}{y}$ equals $-\frac{3}{4}$ at the top and $+\frac{3}{4}$ at the bottom for the same $x=3$ : an expression in $x$ alone could not say both. Substitute for $y$ only if an explicit expression is available and the question asks for it.

4)

FALSE. The condition $y\neq 0$ is not a guarantee; it is precisely the place to examine: on the circle, the points (5, 0) and (-5, 0) have $y=0$ , the formula is undefined there, and these are exactly the two points with a vertical tangent. An undefined fraction does not signal that there is no tangent; it signals a tangent that the slope cannot describe, since a vertical line has no slope.

5)

FALSE. The analogy only holds if you differentiate with respect to the SAME variable as the one that appears. $\frac{d}{dx}(x^{3})=3x^{2}$ because $\frac{dx}{dx}=1$ ; $\frac{d}{dx}(y^{3})=3y^{2}\cdot\frac{dy}{dx}=3y^{2}y'$ because there is no reason for $\frac{dy}{dx}$ to equal 1. Leibniz notation makes the error visible: the forgotten factor is always $\frac{dy}{dx}$ , and it equals 1 only where the curve has slope 1, as on the line through the origin bisecting the first quadrant.

Exercise 9: Problem: the van der Waals gas and its critical point

The ideal gas law predicts no liquefaction. The van der Waals equation does, and the point where the gas turns into a liquid with no visible transition is found by setting TWO derivatives equal to zero at once. It is the finest use of the chapter outside mathematics.

a) The equation is $\left(P+\frac{a}{V^{2}}\right)(V-b)=RT$ . At constant temperature, express $P$ in terms of $V$ , then compute $\frac{dP}{dV}$ and $\frac{d^{2}P}{dV^{2}}$ .
b) The critical point is the one where BOTH derivatives are zero. Show that the critical volume is $V_{c}=3b$ .
c) Deduce the critical temperature $T_{c}$ and the critical pressure $P_{c}$ in terms of $a$ , $b$ and $R$ .
d) For carbon dioxide, $a=3.640$ litres squared atmospheres per mole squared, $b=0.04267$ litre per mole and $R=0.08206$ . Compute $V_{c}$ , $T_{c}$ and $P_{c}$ .
e) Compute the ratio $\frac{P_{c}V_{c}}{RT_{c}}$ and show that it equals $\frac{3}{8}$ for EVERY van der Waals gas. What does this result say about the model?

Show the solution

Answers

a) $P=\frac{RT}{V-b}-\frac{a}{V^{2}}$ ; derivatives $-\frac{RT}{(V-b)^{2}}+\frac{2a}{V^{3}}$ and $\frac{2RT}{(V-b)^{3}}-\frac{6a}{V^{4}}$
b) $V_{c}=3b$
c) $T_{c}=\frac{8a}{27Rb}$ , $P_{c}=\frac{a}{27b^{2}}$
d) $0.128$ L/mol; $308$ K; $74.0$ atm
e) $\frac{3}{8}$ for every gas

a)

Isolate: $P+\frac{a}{V^{2}}=\frac{RT}{V-b}$ , so $P=\frac{RT}{V-b}-\frac{a}{V^{2}}$ , with $T$ treated as a constant. Differentiating with respect to $V$ : $\frac{dP}{dV}=-\frac{RT}{(V-b)^{2}}+\frac{2a}{V^{3}}$ , then $\frac{d^{2}P}{dV^{2}}=\frac{2RT}{(V-b)^{3}}-\frac{6a}{V^{4}}$ . You could also differentiate the implicit form without isolating $P$ ; the result is the same, but isolating is trivial here and there is no reason to go without it.

b)

The first condition gives $\frac{RT}{(V-b)^{2}}=\frac{2a}{V^{3}}$ , that is $RT=\frac{2a(V-b)^{2}}{V^{3}}$ . Substitute into the second condition $\frac{2RT}{(V-b)^{3}}=\frac{6a}{V^{4}}$ : $\frac{2}{(V-b)^{3}}\cdot\frac{2a(V-b)^{2}}{V^{3}}=\frac{6a}{V^{4}}$ , which simplifies to $\frac{4a}{(V-b)V^{3}}=\frac{6a}{V^{4}}$ . The factor $a$ cancels, leaving $4V=6(V-b)$ , so $2V=6b$ and $V_{c}=3b$ . The whole trick is to substitute the first condition into the second rather than solving the two separately.

c)

With $V_{c}=3b$ , we have $V_{c}-b=2b$ , so $RT_{c}=\frac{2a(2b)^{2}}{27b^{3}}=\frac{8ab^{2}}{27b^{3}}=\frac{8a}{27b}$ , hence $T_{c}=\frac{8a}{27Rb}$ . Then $P_{c}=\frac{RT_{c}}{2b}-\frac{a}{9b^{2}}=\frac{8a}{54b^{2}}-\frac{a}{9b^{2}}=\frac{4a}{27b^{2}}-\frac{3a}{27b^{2}}=\frac{a}{27b^{2}}$ . The three critical quantities therefore depend only on the two parameters of the gas, which is the strong prediction of the model.

d)

$V_{c}=3\times 0.04267\approx 0.12801$ litre per mole. $T_{c}=\frac{8\times 3.640}{27\times 0.08206\times 0.04267}\approx\frac{29.12}{0.094540}\approx 308.0$ kelvins, about 34.8 degrees Celsius. $P_{c}=\frac{3.640}{27\times 0.04267^{2}}\approx\frac{3.640}{0.049160}\approx 74.0$ atmospheres. The measured values are 304.1 kelvins and 72.8 atmospheres: the gap is about 1 to 2 percent, remarkable for a two-parameter model. On the figure, it is the point of the critical isotherm where the curve flattens out: the isotherm has a horizontal tangent AND changes concavity there.

e)

$\frac{P_{c}V_{c}}{RT_{c}}=\frac{\frac{a}{27b^{2}}\times 3b}{\frac{8a}{27b}}=\frac{\frac{a}{9b}}{\frac{8a}{27b}}=\frac{a}{9b}\times\frac{27b}{8a}=\frac{27}{72}=\frac{3}{8}=0.375$ . Neither $a$ , nor $b$ , nor the gas considered plays any role: the model predicts the SAME value for every substance. It is a universal prediction, hence easy to refute, and experiment gives rather 0.27 to 0.30 for most real gases. The model therefore captures the right qualitative physics, the very existence of a critical point, but not the fine values: exactly the kind of verdict a derivative computation makes it possible to deliver.

Exercise 10: Problem: the thin lens and the speed of the image

The thin lens equation links the object distance and the image distance without giving one in terms of the other in a convenient form. It is an implicit relation, and it answers a direct question: when the object moves closer, how fast does the image move away?

a) The relation is $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$ , where $f$ is fixed. Differentiate it with respect to $p$ and show that $\frac{dq}{dp}=-\frac{q^{2}}{p^{2}}$ .
b) A lens with focal length $f=20$ centimetres receives an object at $p=60$ centimetres. Compute $q$ , then $\frac{dq}{dp}$ .
c) The object moves toward the lens at 5 centimetres per second. In which direction does the image move, and how fast?
d) The magnification is $m=-\frac{q}{p}$ . Compute $\frac{dm}{dp}$ at the same point, then the rate of change of the magnification. Is the image growing or shrinking?
e) What happens to $\frac{dq}{dp}$ as $p$ approaches $f$ from above, and when $p=2f$ ? Interpret each of the two cases.

Show the solution

Answers

a) $\frac{dq}{dp}=-\frac{q^{2}}{p^{2}}$
b) $q=30$ cm; $-0.25$
c) $+1.25$ cm/s: the image moves away
d) $\frac{dm}{dp}=0.0125$ ; $\frac{dm}{dt}=-0.0625$ per second: the image grows
e) $-\infty$ at the focal point; $-1$ at $p=2f$

a)

Differentiate both sides with respect to $p$ , treating $q$ as a function of $p$ and $f$ as a constant: $-\frac{1}{p^{2}}-\frac{1}{q^{2}}\cdot\frac{dq}{dp}=0$ . Hence $\frac{1}{q^{2}}\cdot\frac{dq}{dp}=-\frac{1}{p^{2}}$ and $\frac{dq}{dp}=-\frac{q^{2}}{p^{2}}$ . The sign is negative whatever the values, which is already physical information: moving the object away always brings the image closer, and conversely.

b)

$\frac{1}{q}=\frac{1}{20}-\frac{1}{60}=\frac{3-1}{60}=\frac{1}{30}$ , so $q=30$ centimetres, as the figure shows. Then $\frac{dq}{dp}=-\frac{30^{2}}{60^{2}}=-\frac{900}{3600}=-0.25$ . In other words, a displacement of the object is passed on to the image four times more weakly, in this particular setting.

c)

“Moving toward the lens” means $\frac{dp}{dt}=-5$ centimetres per second. By the chain rule, $\frac{dq}{dt}=\frac{dq}{dp}\cdot\frac{dp}{dt}=(-0.25)\times(-5)=+1.25$ centimetres per second. The positive sign says that $q$ INCREASES: the image moves away from the lens while the object moves toward it. This is what you observe when focusing a projector, and it is also why focusing becomes more and more delicate in close-up photography.

d)

$m=-\frac{q}{p}$ is a quotient whose two terms both vary: $\frac{dm}{dp}=-\frac{\frac{dq}{dp}\cdot p-q}{p^{2}}=-\frac{(-0.25)(60)-30}{3600}=-\frac{-15-30}{3600}=\frac{45}{3600}=0.0125$ per centimetre. Then $\frac{dm}{dt}=0.0125\times(-5)=-0.0625$ per second. Since $m=-\frac{30}{60}=-0.5$ is negative and becomes even more negative, its ABSOLUTE VALUE grows: the image, already inverted, gets larger. This is consistent with part c: an image farther away is a larger image.

e)

As $p\to f^{+}=20^{+}$ , the relation $\frac{1}{q}=\frac{1}{f}-\frac{1}{p}\to 0^{+}$ gives $q\to+\infty$ , so $\frac{dq}{dp}=-\frac{q^{2}}{p^{2}}\to-\infty$ : the sensitivity becomes infinite, and a micron of displacement of the object sends the image metres away. That is why an object placed at the focal point gives no sharp image. When $p=2f=40$ , we find $q=40$ and $\frac{dq}{dp}=-\frac{1600}{1600}=-1$ : this is the point where object and image move at the same speed in opposite directions, the symmetric configuration with magnification $-1$ , the one used to copy a document at full scale.

Exercise 11: The lemniscate of Bernoulli: tangents and double point

The figure shows the lemniscate with equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$ .

a) Check that the origin and the point $(2, 0)$ are on the curve, then show that $y'=\frac{x\left(2-x^{2}-y^{2}\right)}{y\left(x^{2}+y^{2}+2\right)}$ .
b) Find the points where the tangent is horizontal.
c) Find the points where the tangent is vertical.
d) What happens at the origin? Find the slopes of the two branches that cross there.
e) What is the maximum height of the curve above the $x$ -axis?

Show the solution

Answers

a) $y'=\frac{x(2-x^{2}-y^{2})}{y(x^{2}+y^{2}+2)}$
b) $\left(\pm\sqrt{1.5}, \pm\sqrt{0.5}\right)$
c) $(\pm 2, 0)$
d) Double point: slopes $1$ and $-1$
e) $\sqrt{0.5}\approx 0.707$

a)

At $(0, 0)$ : $0=0$ . At $(2, 0)$ : $16=4\times 4$ ✓. Differentiate: $2(x^{2}+y^{2})(2x+2yy')=4(2x-2yy')$ , that is, dividing by $4$ , $(x^{2}+y^{2})(x+yy')=2x-2yy'$ . Gather the $y'$ terms: $y'\left[y(x^{2}+y^{2})+2y\right]=2x-x(x^{2}+y^{2})$ , hence the announced formula. The square on the left side calls for the chain rule TWICE: on the outer square, then on $y^{2}$ .

b)

Numerator zero and denominator not zero. $x=0$ gives, in the equation, $y^{4}=-4y^{2}$ , so $y=0$ : that is the origin, where the denominator also vanishes, to be ruled out. That leaves $x^{2}+y^{2}=2$ : the equation becomes $4=4(x^{2}-y^{2})$ , that is $x^{2}-y^{2}=1$ . With $x^{2}+y^{2}=2$ : $x^{2}=\frac{3}{2}$ and $y^{2}=\frac{1}{2}$ . Four points: $\left(\pm\sqrt{1.5}, \pm\sqrt{0.5}\right)\approx(\pm 1.225, \pm 0.707)$ , where the denominator equals $\pm 0.707\times 4\neq 0$ ✓.

c)

Denominator zero and numerator not zero: $y=0$ with $x\neq 0$ . The equation becomes $x^{4}=4x^{2}$ , so $x=\pm 2$ . At $(2, 0)$ , the numerator equals $2(2-4)=-4\neq 0$ ✓. Two vertical tangents, at the two ends of the curve.

d)

At the origin, numerator AND denominator vanish: the formula decides nothing; it is a singular point. Near the origin, $(x^{2}+y^{2})^{2}$ is of order $4$ and becomes negligible compared with $4(x^{2}-y^{2})$ , of order $2$ : the curve looks like $x^{2}-y^{2}=0$ , that is, the two lines $y=x$ and $y=-x$ . Two branches cross, with slopes $1$ and $-1$ : it is a DOUBLE POINT, neither a cusp nor a vertical tangent.

e)

The height is greatest where the tangent is horizontal and $y>0$ : $y=\sqrt{0.5}\approx 0.707$ . To be convinced of it without the figure: on the curve, $4y^{2}=4x^{2}-(x^{2}+y^{2})^{2}$ , and the only way to make $y$ extreme is to set its derivative equal to zero, which is what part b) did.

Exercise 12: The curve x to the power y equals y to the power x

For $x>0$ and $y>0$ , we study the curve with equation $x^{y}=y^{x}$ . It contains the line $y=x$ , but not only that: $2^{4}=4^{2}$ .

a) By first taking the logarithm of both sides, show that $y'=\frac{\ln y-\frac{y}{x}}{\ln x-\frac{x}{y}}$ .
b) Compute the slope of the tangent at the point $(2, 4)$ .
c) Compute the slope at the point $(4, 2)$ and explain the link with b).
d) At which point of the line $y=x$ does the formula give $\frac{0}{0}$ ? What happens at that point?
e) Give the y-intercept of the tangent at $(2, 4)$ .

Show the solution

Answers

a) $y\ln x=x\ln y$ , then $y'=\frac{\ln y-\frac{y}{x}}{\ln x-\frac{x}{y}}$
b) $\approx-3.18$
c) $\approx-0.31$ : reciprocal slope by symmetry
d) $(e, e)$ : two branches cross
e) $\approx 10.35$

a)

$x^{y}$ cannot be differentiated as a power, nor as an exponential: both the base and the exponent vary. The logarithm turns everything into products: $y\ln x=x\ln y$ . Differentiate: $y'\ln x+\frac{y}{x}=\ln y+\frac{x}{y}y'$ . Gather: $y'\left(\ln x-\frac{x}{y}\right)=\ln y-\frac{y}{x}$ , hence the formula.

b)

At $(2, 4)$ : $y'=\frac{\ln 4-2}{\ln 2-\frac{1}{2}}\approx\frac{1.3863-2}{0.6931-0.5}=\frac{-0.6137}{0.1931}\approx-3.18$ . The curve falls steeply: when $x$ increases a little beyond $2$ , the corresponding $y$ decreases.

c)

At $(4, 2)$ : $y'=\frac{\ln 2-\frac{1}{2}}{\ln 4-2}\approx-0.315$ , the RECIPROCAL of the previous slope. The equation is symmetric under swapping $x$ and $y$ : the curve is symmetric about $y=x$ , and a reflection in that line turns slopes into their reciprocals. $(2, 4)$ and $(4, 2)$ are images of each other.

d)

On $y=x$ , numerator and denominator both equal $\ln x-1$ : they vanish together at $x=e$ . At the point $(e, e)$ , the line $y=x$ and the curved branch through $(2, 4)$ and $(4, 2)$ CROSS: two different slopes at the same point, $1$ for the line and $-1$ for the curved branch by symmetry, and no single-valued formula can give both. It is the same signature as at the double point of a lemniscate.

e)

Tangent line: $y=4-3.1774(x-2)$ , with y-intercept $4+6.3548\approx 10.35$ . Keeping four decimals in the slope before multiplying by $2$ avoids an error in the hundredths.

Exercise 13: The implicit tangent line to estimate, and concavity to know on which side

Consider the curve $x^{3}+y^{3}=9$ , which passes through the point $(1, 2)$ .

a) Compute $y'$ and its value at $(1, 2)$ .
b) Write the tangent line at $(1, 2)$ and use it to estimate the $y$ -coordinate of the point of the curve with $x$ -coordinate $1.1$ .
c) Compute the actual $y$ -coordinate to the nearest thousandth, and give the error of the estimate.
d) Show that $y''=-\frac{18x}{y^{5}}$ on the curve. Was the estimate in b) too large or too small? Justify without computation.
e) At which point of the curve is the tangent parallel to the line $y=-x$ ?

Show the solution

Answers

a) $y'=-\frac{x^{2}}{y^{2}}$ ; $-\frac{1}{4}$
b) $y\approx 1.975$
c) $1.972$ ; error $0.003$
d) $y''=-0.5625<0$ : estimate too large
e) $\left(\sqrt[3]{4.5}, \sqrt[3]{4.5}\right)\approx(1.651, 1.651)$

a)

$3x^{2}+3y^{2}y'=0$ gives $y'=-\frac{x^{2}}{y^{2}}$ . At $(1, 2)$ : $y'=-\frac{1}{4}$ .

b)

$y=2-\frac{1}{4}(x-1)$ . At $x=1.1$ : $y\approx 2-0.025=1.975$ . We replaced the curve, which cannot be written conveniently, by its tangent line, which can always be written.

c)

$y^{3}=9-1.331=7.669$ , so $y=\sqrt[3]{7.669}\approx 1.972$ . Error: $1.975-1.972=0.003$ , that is $0.15$ percent for a step of $0.1$ .

d)

Differentiate $y'=-x^{2}y^{-2}$ as a product: $y''=-2xy^{-2}+2x^{2}y^{-3}y'$ . Replace $y'$ : $y''=-\frac{2x}{y^{2}}-\frac{2x^{4}}{y^{5}}=-\frac{2x(y^{3}+x^{3})}{y^{5}}$ , and $x^{3}+y^{3}=9$ on the curve: $y''=-\frac{18x}{y^{5}}$ . At $(1, 2)$ : $y''=-\frac{18}{32}=-0.5625<0$ , the curve is concave down, so it lies BELOW its tangent line near the point of contact. The estimate from the tangent line was too large, as c) confirms: $1.975>1.972$ .

e)

Slope $-1$ : $-\frac{x^{2}}{y^{2}}=-1$ , that is $y^{2}=x^{2}$ and $y=\pm x$ . With $y=x$ : $2x^{3}=9$ , $x=\sqrt[3]{4.5}\approx 1.651$ , the point $(1.651, 1.651)$ . With $y=-x$ : $x^{3}-x^{3}=0\neq 9$ , impossible. A single point, on the line $y=x$ , as the symmetry of the equation suggested.

Exercise 14: Problem: the production isoquant and the minimum cost

A company produces $Q=K^{1/3}L^{2/3}$ units per day with $L$ workers and $K$ machine-hours. It targets a fixed output $Q=20$ : the curve $K^{1/3}L^{2/3}=20$ in the $(L, K)$ plane is an ISOQUANT. A worker costs $50$ dollars per day, a machine-hour $4$ dollars. The figure below shows the isoquant.

a) By implicit differentiation with respect to $L$ , show that $\frac{dK}{dL}=-\frac{2K}{L}$ .
b) Check that $(L, K)=(8, 125)$ is on the isoquant and compute $\frac{dK}{dL}$ at that point. Interpret.
c) Use the tangent line to estimate the number of machine-hours needed if the company goes down to $7$ workers, then compare with the exact value.
d) The cost is $C=50L+4K$ . By writing that the isoquant is tangent to a line of constant cost, find the cheapest combination and its cost.
e) Compare with the cost of the combination $(8, 125)$ and with how spending is split between labour and machines.

Show the solution

Answers

a) $\frac{dK}{dL}=-\frac{2K}{L}$
b) $-31.25$ machine-hours per worker
c) $156.25$ against $163.27$
d) $L\approx 10.86$ , $K\approx 67.86$ : $814$ dollars
e) $900$ dollars; two thirds for labour

a)

Take logarithms: $\frac{1}{3}\ln K+\frac{2}{3}\ln L=\ln 20$ . Differentiate with respect to $L$ : $\frac{1}{3}\cdot\frac{K'}{K}+\frac{2}{3}\cdot\frac{1}{L}=0$ , so $K'=-\frac{2K}{L}$ . Without logarithms, differentiating the product leads to the same result with twice as many fractional exponents.

b)

$125^{1/3}\times 8^{2/3}=5\times 4=20$ ✓. $\frac{dK}{dL}=-\frac{2\times 125}{8}=-31.25$ . Interpretation: near this point, removing one worker requires about $31$ more machine-hours to maintain output. This number is called the marginal rate of technical substitution.

c)

With the tangent line: $K\approx 125+(-31.25)(7-8)=156.25$ machine-hours. Exactly, $K=\frac{20^{3}}{L^{2}}=\frac{8000}{49}\approx 163.27$ . The isoquant is convex, above its tangent line, so the estimate is too low by about $7$ hours: substitution becomes more and more expensive as workers are removed.

d)

A line of constant cost $50L+4K=C$ has slope $-\frac{50}{4}=-12.5$ . At minimum cost, the isoquant touches it without crossing it: same slope. $-\frac{2K}{L}=-12.5$ gives $K=6.25L$ . In the equation of the isoquant, $K=\frac{8000}{L^{2}}$ : $6.25L^{3}=8000$ , $L^{3}=1280$ , $L\approx 10.86$ workers and $K\approx 67.86$ machine-hours. Cost: $50\times 10.86+4\times 67.86\approx 543+271=814$ dollars per day.

e)

The combination $(8, 125)$ costs $400+500=900$ dollars, that is $86$ dollars more. At the optimum, labour costs $543$ dollars and machines $271$ : exactly two thirds and one third, the exponents of $L$ and $K$ . This is no coincidence: for output of the form $K^{a}L^{b}$ with $a+b=1$ , the tangency condition $\frac{bK}{aL}=\frac{w}{r}$ is equivalent to $\frac{wL}{rK}=\frac{b}{a}$ .

Exercise 15: Problem: the reflection property of the ellipse

Consider the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ and the point $P(3, 2.4)$ . The figure shows the ellipse, its foci $F_{1}$ and $F_{2}$ , the point $P$ and the segments $PF_{1}$ and $PF_{2}$ .

a) Check that $P$ is on the ellipse and that the foci are $F_{1}(-4, 0)$ and $F_{2}(4, 0)$ , given that $c^{2}=a^{2}-b^{2}$ .
b) Compute the slope of the tangent at $P$ and write its equation.
c) Write the equation of the normal line at $P$ and find its intersection $N$ with the $x$ -axis.
d) Compute $PF_{1}$ and $PF_{2}$ , then compare $\frac{NF_{1}}{NF_{2}}$ with $\frac{PF_{1}}{PF_{2}}$ . What do you conclude about the normal line?
e) Compute the angle between the normal line and each of the segments $PF_{1}$ and $PF_{2}$ . Which physical property of the ellipse have you just proved?

Show the solution

Answers

a) $P$ on the ellipse; $c=4$
b) Slope $-0.45$ ; $9x+20y=75$
c) Normal line with slope $\frac{20}{9}$ ; $N(1.92, 0)$
d) $7.4$ and $2.6$ ; equal ratios: bisector
e) Two angles of $46.8°$ : reflection from one focus to the other

a)

$\frac{9}{25}+\frac{5.76}{9}=0.36+0.64=1$ ✓. $a=5$ , $b=3$ , $c=\sqrt{25-9}=4$ : foci at $(\pm 4, 0)$ .

b)

$\frac{2x}{25}+\frac{2yy'}{9}=0$ gives $y'=-\frac{9x}{25y}$ . At $P$ : $y'=-\frac{27}{60}=-0.45$ . Tangent line: $y=2.4-0.45(x-3)$ , that is $9x+20y=75$ , which the point form $\frac{3x}{25}+\frac{2.4y}{9}=1$ multiplied by $75$ also gives.

c)

The normal line is perpendicular to the tangent: slope $\frac{1}{0.45}=\frac{20}{9}$ . Equation: $y=2.4+\frac{20}{9}(x-3)$ . For $y=0$ : $x=3-\frac{9\times 2.4}{20}=3-1.08=1.92$ . So $N(1.92, 0)$ .

d)

$PF_{1}=\sqrt{7^{2}+2.4^{2}}=\sqrt{54.76}=7.4$ and $PF_{2}=\sqrt{1^{2}+2.4^{2}}=\sqrt{6.76}=2.6$ : their sum is $10=2a$ , the very definition of the ellipse. $NF_{1}=5.92$ and $NF_{2}=2.08$ : $\frac{5.92}{2.08}\approx 2.846$ and $\frac{7.4}{2.6}\approx 2.846$ . By the converse of the angle bisector theorem, line $PN$ is the BISECTOR of the angle $F_{1}PF_{2}$ : the normal line splits the angle between the two focal radii into two equal angles.

e)

A direction vector of the normal pointing inward is $\vec{n}=(-9, -20)$ . With $\overrightarrow{PF_{1}}=(-7, -2.4)$ : $\cos\alpha=\frac{63+48}{\sqrt{481}\times 7.4}\approx 0.6839$ , and with $\overrightarrow{PF_{2}}=(1, -2.4)$ : $\cos\beta=\frac{-9+48}{\sqrt{481}\times 2.6}\approx 0.6839$ . Both angles are about $46.8°$ . A ray coming from $F_{1}$ that reflects at $P$ leaves symmetrically with respect to the normal, so it passes through $F_{2}$ : this is the property that lets a whisper spoken at one focus of an elliptical room be heard at the other focus, and that lets a shock wave be focused on a kidney stone placed at the second focus of a reflector.

Corrected exercises: implicit differentiation (201-SN2-RE)

Course recap

Calculations: implicit derivatives to compute

Exercise 0.1: Implicit differentiation: eleven direct calculations

Part A: the basics (/50)

Exercise 1: The principle: differentiate an equation, not a function

Exercise 2: The five structures that leave a y prime

Exercise 3: The folium of Descartes, a curve that is the graph of no function

Exercise 4: Horizontal and vertical tangents: reading the two zeros

Exercise 5: The implicit second derivative and concavity

Part B: problems and reasoning (/50)

Exercise 6: Two families of curves that always cross at right angles

Exercise 7: Conics and the point form of the tangent

Exercise 8: Five statements to correct

Exercise 9: Problem: the van der Waals gas and its critical point

Exercise 10: Problem: the thin lens and the speed of the image

Part C: the classics (/50)

Exercise 11: The lemniscate of Bernoulli: tangents and double point

Exercise 12: The curve x to the power y equals y to the power x

Exercise 13: The implicit tangent line to estimate, and concavity to know on which side

Exercise 14: Problem: the production isoquant and the minimum cost

Exercise 15: Problem: the reflection property of the ellipse

See also

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