# How do you implicitly differentiate 9=e^y/sin^2x-e^x/cos^2y?

Aug 19, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{y} {\csc}^{2} x \cot x + {e}^{x} {\sec}^{2} y}{{e}^{y} {\csc}^{2} x - 2 {e}^{x} {\sec}^{2} y \tan y}$

#### Explanation:

I'm going to use the fairly well-known trigonometric derivatives:

• $\frac{d}{\mathrm{dx}} \csc x = - \csc x \cot x$
• $\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$

First, I'll rewrite the function:

$9 = {e}^{y} {\csc}^{2} x - {e}^{x} {\sec}^{2} y$

When differentiating, we'll need to first use the product rule.

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

Don't forget that differentiating a function of $y$ with respect to $x$ will cause the chain rule to be in effect.

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

Applying the derivatives from before (and using the chain rule on the derivative of $\sec y$):

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

Rewriting and grouping $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms:

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

Solve for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{y} {\csc}^{2} x \cot x + {e}^{x} {\sec}^{2} y}{{e}^{y} {\csc}^{2} x - 2 {e}^{x} {\sec}^{2} y \tan y}$