3.7 – Implicit Differentiation

1.4 – Parametric Equations

Some functions may be implicitly defined in terms of both $x$ and $y$. For example, a circle around the origin $(0,0)$ with radius 1 may be written as $x^2+y^2=1$. In order to find the rate of change of the circle at $x=0.5$, we must use implicit differentiation.

Differentiate x^2+y^2=1

To differentiate $x^2+y^2=2$ with respect to $x$, we first take the derivative of both sides.

$$\frac{d}{dx}{x}^2+\frac{d}{dx}{y}^2=\frac{d}{dx}2$$

By the chain rule, $\frac{d}{dx}{y}^2=2y\frac{dy}{dx}$ .

$$2x+2y\frac{dy}{dx}=0$$

The last step is to solve for $\frac{dy}{dx}$.

$$\frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}$$

To find the derivative at $x=0.5$:

$$0.5^2+y^2=1\Rightarrow y=\pm 0.866$$

Since this circle crosses each point on the x-axis at two y-values, there are two slopes at $x=0.5$.

$$\frac{dy}{dx}=-\frac{x}{y}{|}_{(0.5,\pm 0.866)}$$ $$=\pm 0.577$$

Implicit differentiation may also be used to find the second derivative (and all higher-order derivatives).

The Second Derivative of x^2+y^2=r^2

We start by finding the first derivative:

$$2x+2y{y}^{\prime }=0$$ $$y^{\prime }=\frac{-2x}{2y}=-\frac{x}{y}$$

We then differentiate $\cancel{2}x+\cancel{2}y{y}^{\prime }=0$.

$$1+y^{\prime }{y}^{\prime }+y{y}^{\prime \prime }=1+(y^{\prime }{)}^2+y{y}^{\prime \prime }=0$$

The first derivative can be substituted in.

$$1+(-\frac{x}{y}{)}^2+y{y}^{\prime \prime }=0$$

$y^{\prime \prime }$ can then be isolated.

$$y^{\prime \prime }=\frac{-\frac{x^2}{y^2}-1}{y}$$ $$=-\frac{x^2}{y^3}-\frac{1}{y}$$ $$=-\frac{(x^2+y^2)}{y^3}$$ $$=-\frac{r^2}{y^3}$$

Examples

1. Find $y^{\prime }$ for $x^{2}y+x{y}^2=6$

Solution

$$[2xy+x^2{y}^{\prime }]+[y^2+x\cdot 2y{y}^{\prime }]=0$$ $$x^2{y}^{\prime }+2xy{y}^{\prime }=-2xy-y^2$$ $$y^{\prime }(x^2+2xy)=-(2xy+y^2)$$ $$y^{\prime }=-\frac{(2xy+y^2)}{2xy+x^2}$$

2. Find $y^{\prime }$ for $y^2=\frac{x-1}{x+1}$

Solution

$$2y{y}^{\prime }=\frac{(x-1{)}^{\prime }(x+1)-(x-1)(x+1{)}^{\prime }}{(x+1{)}^2}$$ $$2y{y}^{\prime }=\frac{(x+1)-(x-1)}{(x+1{)}^2}$$ $$2y{y}^{\prime }=\frac{2}{(x+1{)}^2}$$ $$y^{\prime }=\frac{1}{y(x+1{)}^2}$$

3. Find $y^{\prime }$ for $x=\sin y$

Solution

$$1=\cos (y)y^{\prime }$$ $$y^{\prime }=\frac{1}{\cos (y)}$$

4. Find $y^{\prime \prime }$ for $y^2+2y=2x+1$

Solution

$$2y{y}^{\prime }+2y^{\prime }=2$$ $$y{y}^{\prime }+y^{\prime }=1$$ $$y^{\prime }(y+1)=1$$ $$y^{\prime }=\frac{1}{y+1}$$ $$y^{\prime }{y}^{\prime }+y{y}^{\prime \prime }+y^{\prime \prime }=0$$ $$y^{\prime \prime }(y+1)=-(y^{\prime }{)}^2$$ $$y^{\prime \prime }=-\frac{(y^{\prime }{)}^2}{y+1}$$ $$y^{\prime \prime }=-\frac{(\frac{1}{y+1}{)}^2}{y+1}$$ $$y^{\prime \prime }=-\frac{1}{(y+1{)}^3}$$

5. Find $y^{\prime }$ for $x^2{y}^2-5x^{2}y+x=1$

Solution

$$[(2x)(y^2)+(x^2)(2y{y}^{\prime })]-5[(2x)(y)+(x^2)(y^{\prime })]+1=0$$ $$2x{y}^2+2x^{2}y{y}^{\prime }-10xy-5x^2{y}^{\prime }=-1$$ $$2x^{2}y{y}^{\prime }-5x^2{y}^{\prime }=-1-2x{y}^2+10xy$$ $$y^{\prime }=\frac{10xy-2x{y}^2-1}{2x^{2}y-5x^2}$$

6. Find $y^{\prime }$ for $\sin (x^2{y}^2)=x$

$$\cos (x^2{y}^2)[(2x)(y^2)+(x^2)(2y{y}^{\prime })]=1$$ $$\cos (x^2{y}^2)[2x{y}^2+2x^{2}y{y}^{\prime }]=1$$ $$2x{y}^2+2x^{2}y{y}^{\prime }=\frac{1}{\cos (x^2{y}^2)}$$ $$2x^{2}y{y}^{\prime }=\frac{1}{\cos (x^2{y}^2)}-2x{y}^2$$ $$y^{\prime }=\frac{1}{2x^{2}y\cos (x^2{y}^2)}-\frac{2x{y}^2}{2x^{2}y}$$ $$=\frac{1}{2x^{2}y\cos (x^2{y}^2)}-\frac{y}{x}$$

7. Find $y^{\prime }$ for $xy=8$ in two ways

Solution

Product rule:

$$y+x{y}^{\prime }=8$$ $$x{y}^{\prime }=8-y$$ $$y^{\prime }=\frac{8-y}{x}$$

Isolating $y$:

$$y=\frac{8}{x}$$ $$y^{\prime }=8\frac{d}{dx}{x}^{-1}$$ $$y^{\prime }=-8x^{-2}=-\frac{8}{x^2}$$