Implicit Differentiation

Example 1:

$$ x^3y^2=1,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} x^3y^2 & = 1 \\ \frac{d}{dx} \,x^3y^2 &=\frac{d}{dx}\,1\\ x^3\frac{d}{dx} \, y^2 + y^2\frac{d}{dx} \, x^3 & = 0 \\ x^3\frac{d}{dy} \, y^2 \frac{dy}{dx} + y^2(3x^2) & = 0 \\ x^3(2y) \frac{dy}{dx} + y^2(3x^2) & = 0 \\ 2x^3y \frac{dy}{dx} + 3x^2y^2 & = 0 \\ 2x^3y \frac{dy}{dx} & = -3x^2y^2 \\ \frac{dy}{dx} & = -\frac{3x^2y^2}{2x^3y} \\ \frac{dy}{dx} & = -\frac{3y}{2x} \\ \end{align*}

Example 2:

$$ x^2+y^2=r^2, \;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} x^2+y^2=r^2\\ \frac{d}{dx}x^2+\frac{d}{dx}y^2&=\frac{d}{dx}r^2\\ \frac{d}{dx}x^2+\frac{d}{dy}y^2 \frac{dy}{dx}&=0\\ 2x^{2-1}+2y^{2-1}\frac{dy}{dx}&=0\\ 2x+2y\frac{dy}{dx}&=0\\ 2y\frac{dy}{dx}&=-2x\\ \frac{dy}{dx}&=\frac{-2x}{2y}\\ \frac{dy}{dx}&=-\frac{x}{y}\\ \end{align*}

Example 3:

$$ y=xy+\sin y, \;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=xy+\sin y\\ \frac{dy}{dx}&=\frac{d}{dx}(xy+\sin y)\\ \frac{dy}{dx}&=\frac{d}{dx}xy+\frac{d}{dx} \sin y\\ \frac{dy}{dx}&=x\frac{dy}{dx}+y\frac{d}{dx}x+\frac{d}{dy} \sin y \frac{dy}{dx}\\ \frac{dy}{dx}&=x\frac{dy}{dx}+y(1)+\cos y \frac{dy}{dx}\\ \frac{dy}{dx}&=x\frac{dy}{dx}+y+\cos y \frac{dy}{dx}\\ \frac{dy}{dx}&-x\frac{dy}{dx}-\cos y \frac{dy}{dx}=y\\ (1&-x-\cos y) \frac{dy}{dx}=y\\ \frac{dy}{dx}&=\frac{y}{1-x-\cos y}\\ \end{align*}

Example 4:

$$ \sin (x+y) = \ln (x-y), \;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} \sin (x+y) &= \ln (x-y) \\ \frac{d}{dx} \sin (x+y) &= \frac{d}{dx} \ln (x-y) \\ \cos (x+y) \frac{d}{dx} (x+y) &= \frac{1}{x-y} \frac{d}{dx} (x-y) \\ \cos (x+y) \left(\frac{d}{dx}x+\frac{dy}{dx}\right) &= \frac{1}{x-y} \left(\frac{d}{dx}x - \frac{dy}{dx}\right)\\ \cos (x+y) \left(1+\frac{dy}{dx}\right) &= \frac{1}{x-y} \left(1 - \frac{dy}{dx}\right)\\ (x-y) \cos (x+y) \left(1+\frac{dy}{dx}\right) &= 1 - \frac{dy}{dx}\\ (x-y) \cos (x+y) + (x-y) \cos (x+y) \frac{dy}{dx} &= 1 - \frac{dy}{dx}\\ \frac{dy}{dx} + (x-y) \cos (x+y) \frac{dy}{dx} &= 1 - (x-y) \cos (x+y) \\ \frac{dy}{dx} (1+ (x-y) \cos (x+y))&= 1 - (x-y) \cos (x+y) \\ \frac{dy}{dx} &= \frac{1 - (x-y) \cos (x+y)}{1+ (x-y) \cos (x+y)}\\ \end{align*}