# How do you implicitly differentiate 2=e^(x-y)/tan(y)?

##### 1 Answer
May 4, 2017

Please see the explanation.

#### Explanation:

Given: Implicitly differentiate $2 = {e}^{x - y} / \tan \left(y\right)$

Change $\frac{1}{\tan} \left(y\right) \to \cot \left(y\right)$:

$2 = {e}^{x - y} \cot \left(y\right)$

Differentiate both sides:

$\frac{d \left(2\right)}{\mathrm{dx}} = \frac{d \left({e}^{x - y} \cot \left(y\right)\right)}{\mathrm{dx}}$

The left side is 0:

$0 = \frac{d \left({e}^{x - y} \cot \left(y\right)\right)}{\mathrm{dx}}$

Use the product rule on the right side:

$0 = \frac{d \left({e}^{x - y}\right)}{\mathrm{dx}} \cot \left(y\right) + {e}^{x - y} \frac{d \left(\cot \left(y\right)\right)}{\mathrm{dx}}$

Substitute $- {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$ for $\frac{d \left(\cot \left(y\right)\right)}{\mathrm{dx}}$

$0 = \frac{d \left({e}^{x - y}\right)}{\mathrm{dx}} \cot \left(y\right) + {e}^{x - y} \left(- {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Write ${e}^{x - y}$ as ${e}^{x} {e}^{-} y$:

$0 = \frac{d \left({e}^{x} {e}^{-} y\right)}{\mathrm{dx}} \cot \left(y\right) + {e}^{x - y} \left(- {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Use the product rule:

$0 = \left({e}^{x} {e}^{-} y - {e}^{x} {e}^{-} y \frac{\mathrm{dy}}{\mathrm{dx}}\right) \cot \left(y\right) + {e}^{x - y} \left(- {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Return to the ${e}^{x - y}$ form:

$0 = \left({e}^{x - y} - {e}^{x - y} \frac{\mathrm{dy}}{\mathrm{dx}}\right) \cot \left(y\right) + {e}^{x - y} \left(- {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Use the distributive property:

$0 = {e}^{x - y} \cot \left(y\right) - {e}^{x - y} \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x - y} {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Move the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left:

${e}^{x - y} \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x - y} {\csc}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - y} \cot \left(y\right)$

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\left({e}^{x - y} \cot \left(y\right) + {e}^{x - y} {\csc}^{2} \left(y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - y} \cot \left(y\right)$

${e}^{x - y}$ is on both sides so it cancels:

$\left(\cot \left(y\right) + {\csc}^{2} \left(y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \cot \left(y\right)$

Divide by sides by $\cot \left(y\right) + {\csc}^{2} \left(y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cot \frac{y}{\cot \left(y\right) + {\csc}^{2} \left(y\right)}$