Exercises — Implicit differentiation — technique, applications

$\frac{d}{dx}(x^n)=n{x}^{n-1}$ because the inside ($x$) changes at rate $1$. But $\frac{d}{dx}(y^n)=n{y}^{n-1}\cdot \frac{dy}{dx}$ because the inside ($y$) changes at rate $\frac{dy}{dx}$ as $x$ moves. Same rule, extra factor.

Differentiate both sides w.r.t. $x$. WHAT: apply $\frac{d}{dx}$ to each term. WHY: a balanced equation stays balanced. $2x+2y\frac{dy}{dx}=0$ The $y^2$ term paid the toll: $\frac{d}{dx}(y^2)=2y\cdot y^{\prime }$. Solve: $\frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}.$

Differentiate each term. WHAT: apply $\frac{d}{dx}$ term by term. WHY the toll only on $y^3$: $x^3$ is a plain power of the variable, so $\frac{d}{dx}(x^3)=3x^2$ with no toll; but $y^3$ hides $y$ inside, so the chain rule attaches $\frac{dy}{dx}$: $3x^2+3y^2\frac{dy}{dx}=0.$ WHY the right side is $0$: the constant $10$ never changes, so its derivative is $0$. Now isolate $y^{\prime }$ — WHY divide by $3y^2$: it is the coefficient multiplying $y^{\prime }$: $3y^2\frac{dy}{dx}=-3x^2⟹\frac{dy}{dx}=-\frac{x^2}{y^2}.$

$2y\frac{dy}{dx}=4⟹\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}.$ WHAT IT LOOKS LIKE: at the vertex $y=0$ this blows up — the tangent is vertical there, which matches the picture of a parabola opening rightward.

$xy$ is a product of two functions. WHY product rule: neither factor is constant, so $\frac{d}{dx}(xy)=x^{\prime }y+x{y}^{\prime }$. Recall from the definition above that $x^{\prime }=\frac{dx}{dx}=1$, so this is $1\cdot y+x{y}^{\prime }=y+x\frac{dy}{dx}$. $y+x\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{y}{x}.$

Both $x^{2}y$ and $y^{2}x$ are products (an $x$-piece times a $y$-piece), so each needs the product rule. WHAT + WHY on $x^{2}y$: think "(derivative of $x^2$)$\times y$ + $x^2\times$(derivative of $y$)". The first piece $\frac{d}{dx}(x^2)=2x$ has no toll; the second piece $\frac{d}{dx}(y)=y^{\prime }$ does. Result: $2xy+x^2{y}^{\prime }$. WHAT + WHY on $x{y}^2$: "(derivative of $x$)$\times y^2$ + $x\times$(derivative of $y^2$)" $=(1)y^2+x(2y{y}^{\prime })$ — the $y^2$ pays a toll via the chain rule, the lone $x$ does not. $\underset{\frac{d}{dx}(x^{2}y)}{\underbrace{2xy+x^2{y}^{\prime }}}+\underset{\frac{d}{dx}(x{y}^2)}{\underbrace{y^2+2xy{y}^{\prime }}}=0.$ WHY collect $y^{\prime }$ terms: we want $y^{\prime }$ alone, so gather everything carrying a $y^{\prime }$ on one side and everything else on the other: $y^{\prime }(x^2+2xy)=-(2xy+y^2)⟹\frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}.$

$\cos y\frac{dy}{dx}+1=2y\frac{dy}{dx}.$ WHY $\cos y\cdot y^{\prime }$: $\frac{d}{dx}(\sin y)$ = (derivative of $\sin$ at $y$) $\times$ (rate $y$ changes) $=\cos y\cdot y^{\prime }$ — the chain rule. Collect: $\cos y{y}^{\prime }-2y{y}^{\prime }=-1⟹y^{\prime }(\cos y-2y)=-1⟹\frac{dy}{dx}=\frac{1}{2y-\cos y}.$

$3x^2+3y^2{y}^{\prime }=6(y+x{y}^{\prime }).$ WHY: RHS $6xy$ is a product $\Rightarrow 6(y+x{y}^{\prime })$. Collect: $3y^2{y}^{\prime }-6x{y}^{\prime }=6y-3x^2⟹y^{\prime }=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}.$ At $(3,3)$: $y^{\prime }=\frac{2(3)-9}{9-6}=\frac{-3}{3}=-1.$ The tangent line: $y-3=-1(x-3)\Rightarrow y=-x+6$.

From L1.1, $y^{\prime }=-x/y$.

• Horizontal tangent means $y^{\prime }=0$. A fraction is $0$ when its numerator is $0$: $x=0$. Then $y^2=25\Rightarrow y=\pm 5$. Points: $(0,5)$ and $(0,-5)$ — top and bottom of the circle.
• Vertical tangent means $y^{\prime }$ is undefined, i.e. the denominator is $0$: $y=0$. Then $x=\pm 5$. Points: $(5,0)$ and $(-5,0)$ — left and right edges.

WHAT IT LOOKS LIKE: the four "compass points" of the circle — exactly where the tangent flips between flat and upright.

Differentiate: $2x+(y+x{y}^{\prime })+2y{y}^{\prime }=0$ (product rule on $xy$). $x{y}^{\prime }+2y{y}^{\prime }=-2x-y⟹y^{\prime }(x+2y)=-(2x+y)⟹y^{\prime }=-\frac{2x+y}{x+2y}.$ At $(1,2)$: $y^{\prime }=-\frac{2+2}{1+4}=-\frac{4}{5}.$ Tangent: $y-2=-\frac{4}{5}(x-1)\Rightarrow y=-\frac{4}{5}x+\frac{14}{5}.$

Start from $y^{\prime }=-x/y$. Differentiate again with the quotient rule, remembering $y^{\prime }=-x/y$: $y^{\prime \prime }=-\frac{(1)y-x{y}^{\prime }}{y^2}=-\frac{y-x(-x/y)}{y^2}=-\frac{y+x^2/y}{y^2}=-\frac{y^2+x^2}{y^3}.$ Since $x^2+y^2=25$: $y^{\prime \prime }=-\frac{25}{y^3}.$ At $(3,4)$: $y^{\prime \prime }=-\frac{25}{64}.$ WHY substitute $y^{\prime }$ back: the final answer must not contain $y^{\prime }$ — otherwise it isn't a closed formula for the second derivative.

WHY log first: the base $x$ and the exponent $\sin x$ both vary — neither the power rule nor the exponential rule applies. Taking $\ln$ converts the power into a product. $\ln y=\sin x\cdot \ln x.$ Differentiate implicitly (left side pays a toll, right side needs the product rule): $\frac{1}{y}{y}^{\prime }=\cos x\ln x+\sin x\cdot \frac{1}{x}.$ $y^{\prime }=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).$ At $x=\pi /2$: $\sin x=1,\cos x=0$, so $y=(\pi /2{)}^1=\pi /2$ and $y^{\prime }=\frac{\pi }{2}\left(0+\frac{1}{\pi /2}\right)=\frac{\pi }{2}\cdot \frac{2}{\pi }=1.$

Let $y=\arctan x$, so $\tan y=x$. Differentiate implicitly: ${\sec }^{2}y\frac{dy}{dx}=1⟹\frac{dy}{dx}=\frac{1}{{\sec }^{2}y}.$ Convert back to $x$: ${\sec }^{2}y=1+{\tan }^{2}y=1+x^2$ (Pythagorean identity). Therefore $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.$

Outer chain rule on $\sin$, then product rule on $xy$: $\cos (xy)(y+x{y}^{\prime })=1+y^{\prime }.$ Expand and collect: $y\cos (xy)+x\cos (xy)y^{\prime }=1+y^{\prime }⟹y^{\prime }(x\cos (xy)-1)=1-y\cos (xy).$ $\frac{dy}{dx}=\frac{1-y\cos (xy)}{x\cos (xy)-1}.$ At $(0,0)$: $\cos (0)=1$, so $y^{\prime }=\frac{1-0}{0-1}=-1.$ WHY implicit is the only way: $\sin (xy)=x+y$ cannot be solved for $y$ in closed form.

Both $x$ and $y$ depend on time $t$, so differentiate the equation w.r.t. $t$ (see Related Rates): $2x\frac{dx}{dt}+8y\frac{dy}{dt}=0.$ Plug in $x=2,y=1,dx/dt=3$: $2(2)(3)+8(1)\frac{dy}{dt}=0⟹12+8\frac{dy}{dt}=0⟹\frac{dy}{dt}=-\frac{12}{8}=-\frac{3}{2}.$ So $y$ is decreasing at $1.5$ units/s. WHAT IT LOOKS LIKE: moving rightward on the upper-right of the ellipse means sliding down, so a negative $dy/dt$ is exactly right.

Differentiate: the left side $e^y$ pays a toll. $e^y{y}^{\prime }=1+y^{\prime }⟹y^{\prime }(e^y-1)=1⟹\frac{dy}{dx}=\frac{1}{e^y-1}.$ Check the point: at $(1,0)$, $e^0=1$ and $x+y=1$ ✓. But then $y^{\prime }=\frac{1}{e^0-1}=\frac{1}{0}$ — undefined! WHAT THIS MEANS: the tangent at $(1,0)$ is vertical. Write the equation as $F(x,y)=e^y-x-y=0$; then $F_y=e^y-1$, which is exactly $0$ at $(1,0)$. This is the case the Implicit Function Theorem warns about: where $F_y=0$, the curve cannot be written as a differentiable function $y(x)$. The formula didn't fail — it detected the vertical tangent.