Corrected exercises: implicit differentiation and inverse trigonometric derivatives (MATH 140)

a) Each $y$-term is a composition with inner function $y(x)$. Power and chain rules: $\frac{d}{dx}(y^4) = 4y^3 \cdot y'$. Product rule, then the chain rule on the second factor: $\frac{d}{dx}(x^2y^3) = 2x \cdot y^3 + x^2 \cdot 3y^2y'$. Chain rule with outer function $\cos$: $\frac{d}{dx}(\cos y) = -\sin y \cdot y'$. Chain rule with outer function $e^{u}$, $u = 3y$: $\frac{d}{dx}(e^{3y}) = e^{3y} \cdot 3y'$. Quotient rule: $\frac{d}{dx}\left(\frac{x}{y}\right) = \frac{1 \cdot y - x \cdot y'}{y^2}$. The single rule behind all five: a $y$ is never differentiated to $1$, it is differentiated to $y'$, exactly as $\frac{d}{dx}(x^2 + 1)^4$ carries the factor $2x$. A term like $4y^3$ without its $y'$ is the most expensive slip of the chapter, because it removes the very unknown the question asks for.

b) $(5 - 2)^2 + (3 + 1)^2 = 9 + 16 = 25$: $P$ is on the circle. Differentiate both sides with respect to $x$, chain rule on each square: $2(x - 2) + 2(y + 1)y' = 0$, so $y' = -\frac{x - 2}{y + 1}$ wherever $y \ne -1$. At $P$: $y' = -\frac{3}{4}$. Tangent: $y - 3 = -\frac{3}{4}(x - 5)$, that is $y = -\frac{3}{4}x + \frac{27}{4}$. The radius $CP$ has slope $\frac{3 - (-1)}{5 - 2} = \frac{4}{3}$, and $-\frac{3}{4} \cdot \frac{4}{3} = -1$: the tangent is perpendicular to the radius, as the geometry of the circle demands. That check costs one line and catches a sign error at once.

c) $9 \cdot 1 + 1 = 10$: the point is on the curve. Differentiate, product rule on $x^2y$ and chain rule on $y^3$: $2xy + x^2y' + 3y^2y' = 0$, so $y' = -\frac{2xy}{x^2 + 3y^2}$. At $(3, 1)$: $y' = -\frac{6}{12} = -\frac{1}{2}$, and the tangent is $y - 1 = -\frac{1}{2}(x - 3)$, that is $y = -\frac{1}{2}x + \frac{5}{2}$. For each fixed $x$, the expression $y^3 + x^2y$ is increasing in $y$, so the equation has exactly one real solution $y$: here $y$ IS a function of $x$. But that function is the root of a cubic, whose formula is unusable by hand; implicit differentiation gives the slope without ever writing it. A faster route with numbers: substitute $x = 3$, $y = 1$ into $2xy + (x^2 + 3y^2)y' = 0$ before isolating, which gives $6 + 12y' = 0$ directly.

d) With the wrong derivative $2xy'$, the equation becomes $2xy' + 3y^2y' = 0$, that is $y'(2x + 3y^2) = 0$. At $(3, 1)$, $2x + 3y^2 = 9 \ne 0$, so $y' = 0$, and the same happens at every point where $2x + 3y^2 \ne 0$: the student has proved that the curve is flat. It is not: $(1, 2)$ is also on the curve, since $1 \cdot 2 + 8 = 10$, and the secant from $(1, 2)$ to $(3, 1)$ has slope $\frac{1 - 2}{3 - 1} = -\frac{1}{2}$. The product rule was skipped: $x^2y$ is a product of TWO functions of $x$, and its derivative is $2xy + x^2y'$. An answer $y' = 0$ everywhere on a curve that visibly goes down is the signal to recheck each product.

e) $e^1 + 0 = e$: the point is on the curve. Differentiate: chain rule on $e^y$, product rule on $xy$: $e^y y' + y + xy' = 0$, so $y' = -\frac{y}{e^y + x}$. At $(0, 1)$: $y' = -\frac{1}{e}$, an exact answer; with $e \approx 2.72$ it is about $-0.37$, but the exact form is the one expected. Tangent: $y = 1 - \frac{x}{e}$. Here solving for $y$ is impossible with the functions of the course: $e^y + xy = e$ mixes $y$ inside and outside an exponential. That is the everyday reason implicit differentiation exists.

a) At $(2, 4)$: $2^3 + 4^3 = 8 + 64 = 72$ and $9 \cdot 2 \cdot 4 = 72$. At $(4, 2)$ the same two numbers appear in the other order. Differentiate both sides with respect to $x$: chain rule on $y^3$, product rule on $9xy$: $3x^2 + 3y^2y' = 9y + 9xy'$. Gather the $y'$ terms on one side: $y'(3y^2 - 9x) = 9y - 3x^2$, and divide by $3$: $y' = \frac{3y - x^2}{y^2 - 3x}$, wherever $y^2 \ne 3x$. The gathering step is where signs are lost: move every $y'$ term to the left BEFORE factoring, and factor $y'$ out as a whole.

b) At $(2, 4)$: $y' = \frac{12 - 4}{16 - 6} = \frac{8}{10} = \frac{4}{5}$, tangent $y - 4 = \frac{4}{5}(x - 2)$, that is $y = \frac{4}{5}x + \frac{12}{5}$. At $(4, 2)$: $y' = \frac{6 - 16}{4 - 12} = \frac{-10}{-8} = \frac{5}{4}$, tangent $y - 2 = \frac{5}{4}(x - 4)$, that is $y = \frac{5}{4}x - 3$. Both slopes are positive, which the figure confirms: at $(2, 4)$ the loop is still climbing toward its top, at $(4, 2)$ it climbs steeply toward its right end.

c) Replacing $(x, y)$ by $(y, x)$ turns $x^3 + y^3 = 9xy$ into $y^3 + x^3 = 9yx$, the same equation: the point $(a, b)$ is on the folium exactly when $(b, a)$ is, which is the symmetry in the line $y = x$. That reflection exchanges the roles of $\Delta x$ and $\Delta y$, so a line of slope $m$ becomes a line of slope $\frac{1}{m}$: the tangent at $(4, 2)$ is the reflection of the tangent at $(2, 4)$, and $\frac{5}{4} = \frac{1}{4/5}$. Check on the equations: reflecting $y = \frac{4}{5}x + \frac{12}{5}$ means swapping $x$ and $y$, $x = \frac{4}{5}y + \frac{12}{5}$, that is $y = \frac{5}{4}x - 3$, the second tangent exactly.

d) A horizontal tangent needs the numerator to vanish, $3y - x^2 = 0$, at a point of the curve where the denominator does not. Substitute $y = \frac{x^2}{3}$ into the equation: $x^3 + \frac{x^6}{27} = 9x \cdot \frac{x^2}{3} = 3x^3$, so $\frac{x^6}{27} = 2x^3$. Either $x = 0$, which gives the origin, where the denominator also vanishes and the formula reads $\frac{0}{0}$; or $x^3 = 54$, $x = \sqrt[3]{54} = 3\sqrt[3]{2}$, and $y = \frac{x^2}{3} = \frac{9\sqrt[3]{4}}{3} = 3\sqrt[3]{4}$. Denominator there: $y^2 - 3x = 9\sqrt[3]{16} - 9\sqrt[3]{2} = 18\sqrt[3]{2} - 9\sqrt[3]{2} = 9\sqrt[3]{2} \ne 0$. So the tangent is horizontal at $(3\sqrt[3]{2}, 3\sqrt[3]{4})$, the top of the loop. By the symmetry of c), reflecting a horizontal tangent gives a vertical one: the tangent is vertical at $(3\sqrt[3]{4}, 3\sqrt[3]{2})$. Writing $x^3 = 54$ and stopping at the decimal $3.78$ is not the exact answer the course expects.

e) With $x = 2$: $8 + y^3 = 18y$, that is $y^3 - 18y + 8 = 0$. We know the root $y = 4$ from a), so divide by $y - 4$: $y^3 - 18y + 8 = (y - 4)(y^2 + 4y - 2)$, and the quadratic gives $y = -2 \pm \sqrt 6$. Three points: $(2, 4)$ on the top of the loop, $(2, -2 + \sqrt 6)$ on its lower part, and $(2, -2 - \sqrt 6)$ on the branch in the fourth quadrant. At $(2, -2 + \sqrt 6)$: $3y - 4 = 3\sqrt 6 - 10$ and $y^2 - 6 = (10 - 4\sqrt 6) - 6 = 4 - 4\sqrt 6$, so $y' = \frac{3\sqrt 6 - 10}{4 - 4\sqrt 6} = \frac{7\sqrt 6 - 8}{20}$ after multiplying top and bottom by $4\sqrt 6 + 4$. At $(2, -2 - \sqrt 6)$ the same steps give $y' = -\frac{7\sqrt 6 + 8}{20}$, about $-1.26$, close to the slope $-1$ of the asymptote, as it should be. One value of $x$, three points, three slopes: a formula in $x$ alone would have to return three answers at once. That is why $\frac{dy}{dx}$ must contain $y$, and why every slope question must say at WHICH point.

a) Differentiate, product rule on $xy$: $2x - (y + xy') + 2yy' = 0$, so $y'(2y - x) = y - 2x$ and $y' = \frac{y - 2x}{2y - x}$. Horizontal: $y - 2x = 0$, so $y = 2x$; this is a line, not an answer, and it must be intersected with the curve: $x^2 - 2x^2 + 4x^2 = 3x^2 = 3$, so $x = \pm 1$ and the points $(1, 2)$ and $(-1, -2)$. Denominators: $2 \cdot 2 - 1 = 3$ and $-4 + 1 = -3$, both nonzero, so both tangents are horizontal. Vertical: $2y - x = 0$, so $x = 2y$, and $4y^2 - 2y^2 + y^2 = 3y^2 = 3$: the points $(2, 1)$ and $(-2, -1)$, where the numerator is $1 - 4 = -3$ and $-1 + 4 = 3$, nonzero. Four points in total, the top, bottom, right and left ends of the tilted ellipse.

b) Differentiate, chain rule on $y^2$: $2yy' = 3x^2 + 6x$, so $y' = \frac{3x^2 + 6x}{2y}$. Horizontal: $3x(x + 2) = 0$. For $x = -2$: $y^2 = -8 + 12 = 4$, so $(-2, 2)$ and $(-2, -2)$, where $D = \pm 4 \ne 0$: horizontal tangents, the top and bottom of the loop on the figure. For $x = 0$: $y = 0$, but there $D = 0$ as well, so the origin is NOT a horizontal-tangent point. Vertical: $2y = 0$, so $y = 0$ and $x^2(x + 3) = 0$, that is $x = 0$ or $x = -3$. At $(-3, 0)$: $N = 27 - 18 = 9 \ne 0$, so the tangent is vertical, the left end of the loop. At the origin, $N = 0$ again: excluded.

c) Since $y^2 = x^2(x + 3)$ and $x + 3 \ge 0$ near $0$, $y = \pm\sqrt{x^2(x + 3)} = \pm|x|\sqrt{x + 3}$, and the two branches $y = x\sqrt{x + 3}$ and $y = -x\sqrt{x + 3}$ together give the same set of points. Differentiate the first, product and chain rules: $\sqrt{x + 3} + \frac{x}{2\sqrt{x + 3}}$, which equals $\sqrt 3$ at $x = 0$. The second branch has slope $-\sqrt 3$. Two branches pass through the origin with two different slopes: the curve crosses itself there, as the figure shows. There is no single tangent line, so no single value of $y'$, and the $\frac{0}{0}$ of the formula was the honest answer: the formula cannot choose between two slopes.

d) The student used only half of the rule. $D = 0$ gives a vertical tangent only when $N \ne 0$; at the origin $N = 0$ too, and c) shows two branches of slopes $\pm\sqrt 3$, none of them vertical. The complete rule: at a point of the curve, the tangent is horizontal when the numerator of $y'$ is $0$ and the denominator is not, vertical when the denominator is $0$ and the numerator is not, and when both are $0$ the formula decides nothing and the curve must be studied near the point. On an exam, the sentence 'both vanish, so no conclusion from $y'$' earns the mark that the vertical tangent loses.

e) Parallel to $y = x$ means $y' = 1$: $\frac{y - 2x}{2y - x} = 1$, so $y - 2x = 2y - x$, that is $y = -x$ (with $2y - x \ne 0$). Intersect with the ellipse: $x^2 + x^2 + x^2 = 3x^2 = 3$, so $x = \pm 1$: the points $(1, -1)$ and $(-1, 1)$, where $2y - x = -3$ and $3$, nonzero. Check at $(1, -1)$: $1 + 1 + 1 = 3$ and $y' = \frac{-1 - 2}{-2 - 1} = 1$. Same gesture as a): the condition on $y'$ gives a line, and the curve gives the points.

a) Differentiate: $8x + 18yy' = 0$, so $y' = -\frac{4x}{9y}$. Now differentiate this quotient, $y$ still a function of $x$: $y'' = -\frac{4}{9} \cdot \frac{1 \cdot y - x \cdot y'}{y^2}$. Replace $y'$: $y - xy' = y + \frac{4x^2}{9y} = \frac{9y^2 + 4x^2}{9y}$. Hence $y'' = -\frac{4}{9} \cdot \frac{4x^2 + 9y^2}{9y^3}$. The numerator is the left side of the equation of the ellipse, so $4x^2 + 9y^2 = 36$ ON the curve, and $y'' = -\frac{4 \cdot 36}{81y^3} = -\frac{16}{9y^3}$. This last substitution is legitimate only because we evaluate at points of the ellipse, and it is the step the marker looks for: an answer left as $-\frac{4(4x^2 + 9y^2)}{81y^3}$ is correct but loses the simplification mark.

b) At $(0, 2)$: $y' = 0$ and $y'' = -\frac{16}{9 \cdot 8} = -\frac{2}{9}$. At $(0, -2)$: $y' = 0$ and $y'' = -\frac{16}{9 \cdot (-8)} = \frac{2}{9}$, the opposite sign, because the bottom of the ellipse bends the other way. For the third point: $4 \cdot \frac{9}{4} + 9 \cdot 3 = 9 + 27 = 36$, so it is on the ellipse. There $y' = -\frac{4 \cdot 3/2}{9\sqrt 3} = -\frac{6}{9\sqrt 3} = -\frac{2}{3\sqrt 3} = -\frac{2\sqrt 3}{9}$, and $y'' = -\frac{16}{9 \cdot 3\sqrt 3} = -\frac{16}{27\sqrt 3} = -\frac{16\sqrt 3}{81}$, using $(\sqrt 3)^3 = 3\sqrt 3$.

c) $(1, 1)$ is on the curve: $1 + 1 = 2$. Differentiate: $3x^2 + 3y^2y' = 0$, so $y' = -\frac{x^2}{y^2}$. Then, quotient rule with $y$ still a function: $y'' = -\frac{2x \cdot y^2 - x^2 \cdot 2yy'}{y^4}$. Replace $y'$: $2xy^2 - 2x^2y\left(-\frac{x^2}{y^2}\right) = 2xy^2 + \frac{2x^4}{y} = \frac{2x(y^3 + x^3)}{y}$. So $y'' = -\frac{2x(x^3 + y^3)}{y^5} = -\frac{4x}{y^5}$, because $x^3 + y^3 = 2$ on the curve. At $(1, 1)$: $y' = -1$ and $y'' = -4$. Once more the equation of the curve appears in the numerator, and it is no accident: the numerator was built from the derivative of the same equation.

d) Treating $y$ as a constant, the student writes $y'' = -\frac{4}{9y}$. At $\left(\frac{3}{2}, \sqrt 3\right)$ this gives $-\frac{4}{9\sqrt 3} = -\frac{4\sqrt 3}{27}$, while the true value is $-\frac{16\sqrt 3}{81}$. The ratio is $\frac{4/27}{16/81} = \frac{4 \cdot 81}{27 \cdot 16} = \frac{3}{4}$: the wrong answer is $\frac{3}{4}$ of the right one. The missing quarter is exactly the term $-x \cdot y'$ of the quotient rule, the one that comes from $y$ being a function. Note that at $(0, 2)$ the two answers coincide, because $x = 0$ kills that term: a check at a point with $x = 0$ would not catch the error. Test a formula at a GENERIC point.