Corrected exercises: implicit differentiation (MATH 203)

a) Rewrite first, then differentiate: $\frac{1}{y^2} = y^{-2}$, whose derivative is $-2y^{-3} \cdot y' = -\frac{2y'}{y^3}$ (power rule, chain rule with inner function $y$). Product rule: $\frac{d}{dx}(x^3y^2) = 3x^2y^2 + x^3 \cdot 2yy'$. With $\sqrt y = y^{1/2}$: $\frac{1}{2}y^{-1/2}y' = \frac{y'}{2\sqrt y}$. Chain rule with inner function $u = xy$, whose derivative is $y + xy'$ by the product rule: $\frac{d}{dx}\sin(xy) = \cos(xy)(y + xy')$. Chain rule with inner function $u = x^2 + y$: $\frac{d}{dx}(x^2 + y)^3 = 3(x^2 + y)^2(2x + y')$. With $x = 1$, $y = 2$, $y' = -1$: $-\frac{2(-1)}{8} = \frac{1}{4}$; $3 \cdot 4 + 2 \cdot 2 \cdot (-1) = 8$; $\frac{-1}{2\sqrt 2} = -\frac{\sqrt 2}{4} \approx -0.3536$; $\cos 2 \cdot (2 - 1) = \cos 2 \approx -0.4161$ (radians); $3 \cdot 9 \cdot (2 - 1) = 27$. The algebra that costs marks here is the first rewrite: $\frac{1}{y^2}$ is $y^{-2}$, with the exponent $-2$, and its derivative has the exponent $-3$, not $-1$.

b) $2 + 2 + 4 = 8$: $P$ is on the curve. Differentiate both sides, product rule on $xy$ and chain rule on $y^2$: $4x + (y + xy') + 2yy' = 0$. Collect the $y'$ terms on the left and everything else on the right, CHANGING THE SIGN of what crosses: $xy' + 2yy' = -4x - y$. Factor $y'$ out: $y'(x + 2y) = -(4x + y)$, so $y' = -\frac{4x + y}{x + 2y}$ wherever $x + 2y \ne 0$. At $P$: $y' = -\frac{4 + 2}{1 + 4} = -\frac{6}{5}$. Tangent: $y - 2 = -\frac{6}{5}(x - 1)$, that is $y = -\frac{6}{5}x + \frac{16}{5}$.

c) The normal is perpendicular to the tangent, so its slope is the negative reciprocal $-\frac{1}{-6/5} = \frac{5}{6}$. Normal: $y - 2 = \frac{5}{6}(x - 1)$, that is $y = \frac{5}{6}x + \frac{7}{6}$. Check: $-\frac{6}{5} \cdot \frac{5}{6} = -1$. Writing $-\frac{5}{6}$ (reciprocal without the sign change) or $\frac{6}{5}$ (sign change without the reciprocal) are the two usual slips; the product $-1$ catches both.

d) The first line is right. In the second, $4x + y$ moved to the right side without its sign changing: the correct line is $xy' + 2yy' = -4x - y$. With his line, $y' = \frac{4x + y}{x + 2y}$, which gives $+\frac{6}{5}$ at $P$. The figure refutes it in one glance: at $P$ the curve is going DOWN from left to right, and so is its tangent; a positive slope is impossible. The top of the curve, where $4x + y = 0$, lies at $x = -\sqrt{\frac{4}{7}} \approx -0.76$, to the LEFT of $P$, so the arc through $P$ is past its highest point. A sign error in the collecting step does not show in the calculus; it shows only against the picture or the geometry.

e) $\sin \pi + 4 \cdot 1 = 0 + 4 = 4$: the point is on the curve. Differentiate: chain rule on $\sin(\pi y)$, whose inner function $\pi y$ has derivative $\pi y'$, and product rule on $x^2y$: $\pi\cos(\pi y)\,y' + 2xy + x^2y' = 0$. Factor: $y'(\pi\cos(\pi y) + x^2) = -2xy$. At $(2, 1)$, $\cos \pi = -1$: $y'(4 - \pi) = -4$, so $y' = -\frac{4}{4 - \pi}$, about $-4.6598$ with the calculator in RADIAN mode. Two classic losses: forgetting the factor $\pi$ from the inner function, which gives $-\frac{4}{3}$, and a calculator in degree mode, which reads $\cos\pi$ as the cosine of $3.14$ degrees.

a) Check: $4 + 6 = 10 = 2 + 8$. Differentiate: the term $x^2y^2$ is a product of two functions of $x$, $x^2$ and $y^2$, so the product rule applies, with the chain rule on $y^2$: $2xy^2 + x^2 \cdot 2yy' + 3y' = 2$. Substitute $x = 1$, $y = 2$ at once: $8 + 4y' + 3y' = 2$, so $7y' = -6$ and $y' = -\frac{6}{7}$. The trap is to differentiate $x^2y^2$ as $2x \cdot 2yy'$, the product of the derivatives: that gives $8y' + 3y' = 2$ at the point, and $y' = \frac{2}{11}$, with the wrong sign: the term $2xy^2 = 8$ has vanished.

b) Check: $3\sqrt 4 + 4 = 6 + 4 = 10$. Product rule on $x\sqrt y$, with $\frac{d}{dx}\sqrt y = \frac{y'}{2\sqrt y}$: $\sqrt y + \frac{xy'}{2\sqrt y} + y' = 0$. At $(3, 4)$: $2 + \frac{3y'}{4} + y' = 0$, so $\frac{7}{4}y' = -2$ and $y' = -\frac{8}{7}$. For the general formula, CLEAR THE FRACTION before solving: multiply every term by $2\sqrt y$, which gives $2y + xy' + 2\sqrt y\,y' = 0$, since $\sqrt y \cdot 2\sqrt y = 2y$. Then $y'(x + 2\sqrt y) = -2y$ and $y' = -\frac{2y}{x + 2\sqrt y}$; at $(3, 4)$ this is $-\frac{8}{3 + 4} = -\frac{8}{7}$, as before. Solving without clearing leaves $y' = \frac{-\sqrt y}{\frac{x}{2\sqrt y} + 1}$, a fraction inside a fraction, where a slip is almost certain.

c) Multiply both sides by the common denominator $2xy$ (with $x \ne 0$, $y \ne 0$): $2x^2 + 2y^2 = 5xy$. Differentiate, product rule on $xy$: $4x + 4yy' = 5y + 5xy'$. Collect: $4yy' - 5xy' = 5y - 4x$, so $y' = \frac{5y - 4x}{4y - 5x}$. At $(1, 2)$: check $\frac{1}{2} + 2 = \frac{5}{2}$; $y' = \frac{10 - 4}{8 - 5} = 2$. At $(2, 1)$: $y' = \frac{5 - 8}{4 - 10} = \frac{-3}{-6} = \frac{1}{2}$. Clearing the denominators is legitimate because $2xy \ne 0$ on the curve: the cleared equation describes the same points.

d) $\frac{d}{dx}\frac{x}{y} = \frac{y - xy'}{y^2}$ and $\frac{d}{dx}\frac{y}{x} = \frac{xy' - y}{x^2}$ (quotient rule twice). At $(1, 2)$: $\frac{2 - y'}{4} + \frac{y' - 2}{1} = 0$. Multiply by $4$: $2 - y' + 4y' - 8 = 0$, so $3y' = 6$ and $y' = 2$. Same answer, but two quotient rules, two squares in the denominators, and one more fraction to clear at the end. In general the quotient rule on a $y$ in a denominator is the most error-prone line of the chapter: when the equation allows it, multiply out first.

e) $2x^2 - 5xy + 2y^2 = (2x - y)(x - 2y)$, as expanding confirms: $2x^2 - 4xy - xy + 2y^2$. So the curve is the pair of lines $y = 2x$ and $y = \frac{x}{2}$, without the origin, where the original fractions are undefined. $(1, 2)$ lies on $y = 2x$, of slope $2$; $(2, 1)$ lies on $y = \frac{x}{2}$, of slope $\frac{1}{2}$. The formula agrees: on $y = 2x$, $\frac{5y - 4x}{4y - 5x} = \frac{10x - 4x}{8x - 5x} = 2$ at every point. Factoring before differentiating would have given the slopes with no calculus at all: always look at the cleared equation before launching the computation.

a) Differentiate, product rule on $2xy$ and chain rule on $2y^2$: $2x + 2y + 2xy' + 4yy' = 0$. Collect and factor: $y'(2x + 4y) = -(2x + 2y)$, then divide by the common factor $2$: $y' = -\frac{x + y}{x + 2y}$. At $(0, \sqrt 2)$, on the ellipse since $2 \cdot 2 = 4$: $y' = -\frac{\sqrt 2}{2\sqrt 2} = -\frac{1}{2}$. Simplifying by the common factor $2$ is not cosmetic: the conditions of b) and c) read directly on the simplified form.

b) $N = 0$ means $x + y = 0$, the line $y = -x$. Substitute into the ellipse: $x^2 - 2x^2 + 2x^2 = x^2 = 4$, so $x = \pm 2$: the points $(2, -2)$ and $(-2, 2)$. Check $D = x + 2y$: $2 - 4 = -2$ and $-2 + 4 = 2$, both nonzero. Two horizontal tangents, the lowest and the highest points of the figure. Stopping at 'the tangent is horizontal when $y = -x$' answers nothing: the line crosses the plane, the ellipse only twice.

c) $D = 0$ means $x = -2y$. Substitute: $4y^2 - 4y^2 + 2y^2 = 2y^2 = 4$, so $y = \pm\sqrt 2$ and $x = \mp 2\sqrt 2$: the points $(-2\sqrt 2, \sqrt 2)$ and $(2\sqrt 2, -\sqrt 2)$. There $N = x + y = -\sqrt 2$ and $\sqrt 2$, nonzero: vertical tangents, the leftmost and rightmost points, at $x \approx \pm 2.83$ on the figure. The substitution is where the algebra bites: $(-2y)^2 = 4y^2$, and $2(-2y)y = -4y^2$, two signs to get right.

d) Parallel to $x + y = 0$ means slope $-1$: $-\frac{x + y}{x + 2y} = -1$, so $x + y = x + 2y$, that is $y = 0$ (with $x + 2y \ne 0$). On the ellipse, $y = 0$ gives $x^2 = 4$: the points $(2, 0)$ and $(-2, 0)$, where $D = \pm 2 \ne 0$. Cross-multiplying is safe here only because $D \ne 0$ at the points kept.

e) The tangent at a point $(x_0, y_0)$ of the ellipse passes through the origin exactly when the slope of the segment from $O$ equals the slope of the tangent: $\frac{y_0}{x_0} = -\frac{x_0 + y_0}{x_0 + 2y_0}$ (the case $x_0 = 0$ is separate, see below). Cross-multiply: $x_0y_0 + 2y_0^2 = -x_0^2 - x_0y_0$, that is $x_0^2 + 2x_0y_0 + 2y_0^2 = 0$. But the point is on the ellipse, where this expression equals $4$: contradiction, so no such point. If $x_0 = 0$, then $y_0 = \pm\sqrt 2$ and the tangents have slope $-\frac{1}{2}$ by a), so they are $y = \pm\sqrt 2 - \frac{x}{2}$, which miss the origin. The equation of the curve does the whole proof: at a point of the curve, the coordinates are not free, they satisfy the equation.

a) Differentiate: $6x - 2yy' = 0$, so $y' = \frac{3x}{y}$. Quotient rule, $y$ still a function of $x$: $y'' = \frac{3 \cdot y - 3x \cdot y'}{y^2}$. Replace $y'$: $y'' = \frac{3y - \frac{9x^2}{y}}{y^2}$, a complex fraction. Multiply the numerator AND the denominator by $y$: $y'' = \frac{3y^2 - 9x^2}{y^3} = -\frac{3(3x^2 - y^2)}{y^3}$. On the curve $3x^2 - y^2 = 11$, so $y'' = -\frac{33}{y^3}$. Three algebraic gestures in a row, each worth a mark: clear the complex fraction, factor out $-3$, recognise the equation of the curve.

b) $12 - 1 = 11$ at both points. At $(2, 1)$: $y' = \frac{6}{1} = 6$ and $y'' = -\frac{33}{1} = -33$. At $(2, -1)$: $y' = \frac{6}{-1} = -6$ and $y'' = -\frac{33}{-1} = 33$. The two points are symmetric in the $x$-axis, so the slopes are opposite and so is $y''$; the formula $-\frac{33}{y^3}$ carries that symmetry in the odd power $y^3$.

c) Check: $1 + 9 = 10$. Differentiate, chain rule on $y^3$ and product rule on $xy$: $3y^2y' + y + xy' = 0$. At $(9, 1)$: $3y' + 1 + 9y' = 0$, so $y' = -\frac{1}{12}$. Differentiate the equation $3y^2y' + y + xy' = 0$ once more, $y$ and $y'$ both functions of $x$: product rule on $3y^2 \cdot y'$ gives $6y(y')^2 + 3y^2y''$, the term $y$ gives $y'$, and product rule on $xy'$ gives $y' + xy''$. So $6y(y')^2 + 3y^2y'' + 2y' + xy'' = 0$. At $(9, 1)$ with $y' = -\frac{1}{12}$: $\frac{6}{144} + 3y'' - \frac{2}{12} + 9y'' = 0$, that is $12y'' = \frac{1}{6} - \frac{1}{24} = \frac{1}{8}$, and $y'' = \frac{1}{96}$. The numbers go in only AFTER the second differentiation; the term $6y(y')^2$ is the one that disappears if $y'$ is frozen too early.

d) He multiplied only the numerator by $y$: $3y \cdot y = 3y^2$ and $\frac{9x^2}{y} \cdot y = 9x^2$, but left the denominator $y^2$ instead of $y^3$. Multiplying one floor of a fraction changes its value; the correct move multiplies both floors. At $(2, 1)$ his formula gives $\frac{3 - 36}{1} = -33$, the right value, because there $y = 1$ and the forgotten factor $y$ equals $1$. At $(3, 4)$, on the curve since $27 - 16 = 11$: his formula gives $\frac{48 - 81}{16} = -\frac{33}{16}$, while the truth is $-\frac{33}{64}$. He is off by a factor $4$, which is exactly $y$. A formula tested at a point where $y = 1$ tests nothing about powers of $y$.

e) Differentiate, product rule: $y'(x + 2) + y = 0$, so $y' = -\frac{y}{x + 2}$. Quotient rule: $y'' = -\frac{y'(x + 2) - y}{(x + 2)^2}$. Replace $y'(x + 2)$ by $-y$, which the first line gives directly: $y'' = -\frac{-y - y}{(x + 2)^2} = \frac{2y}{(x + 2)^2}$. With the equation, $y = \frac{4}{x + 2}$, so $y'' = \frac{8}{(x + 2)^3}$. Explicitly, $y = 4(x + 2)^{-1}$, $y' = -4(x + 2)^{-2}$ and $y'' = 8(x + 2)^{-3}$: the same. At $x = 0$: $y'' = \frac{8}{8} = 1$. Substituting the whole product $y'(x + 2)$ instead of $y'$ alone saved a fraction: look for the block the first line already gives.