# How do you implicitly differentiate 9=e^ysin^2x-e^xcos^2y?

$y ' = \frac{{e}^{x} {\cos}^{2} y - 2 \cdot {e}^{y} \sin x \cos x}{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}$

#### Explanation:

We start from the given equation

$9 = {e}^{y} {\sin}^{2} x - {e}^{x} {\cos}^{2} y$

Differentiate with respect to x both sides of the equation

$9 = {e}^{y} {\sin}^{2} x - {e}^{x} {\cos}^{2} y$

$\frac{d}{\mathrm{dx}} \left(9\right) = \frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x\right) - \frac{d}{\mathrm{dx}} \left({e}^{x} {\cos}^{2} y\right)$

Use the product formula $\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{d}{\mathrm{dx}} \left(u\right) + u \frac{d}{\mathrm{dx}} \left(v\right)$ for differentiating the right side of the equation

$0 = {e}^{y} \cdot \frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) + {\sin}^{2} x \cdot \frac{d}{\mathrm{dx}} \left({e}^{y}\right) - \left[{e}^{x} \cdot \frac{d}{\mathrm{dx}} \left({\cos}^{2} y\right) + {\cos}^{2} y \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)\right]$

$0 = {e}^{y} \cdot 2 \cdot \sin x \cos x + {\sin}^{2} x \cdot \left({e}^{y}\right) y ' - \left[{e}^{x} \cdot 2 \left({\cos}^{2 - 1} y\right) \left(- \sin y\right) y ' + {\cos}^{2} y \cdot \left({e}^{x}\right) \cdot 1\right]$

Simplify

$0 = 2 {e}^{y} \sin x \cos x + {e}^{y} {\sin}^{2} x \cdot y ' + 2 {e}^{x} \sin y \cos y y ' - {e}^{x} {\cos}^{2} y$

Solve now for $y '$

Start transposing the terms with $y '$ to one side then factor out $y '$

${e}^{y} {\sin}^{2} x \cdot y ' + 2 {e}^{x} \sin y \cos y y ' = {e}^{x} {\cos}^{2} y - 2 {e}^{y} \sin x \cos x$

$\left({e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y\right) y ' = {e}^{x} {\cos}^{2} y - 2 {e}^{y} \sin x \cos x$

Divide both sides of the equation by the coefficient of $y '$

$\frac{\left({e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y\right) y '}{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y} = \frac{{e}^{x} {\cos}^{2} y - 2 {e}^{y} \sin x \cos x}{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}$

$\frac{\left(\cancel{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}\right) y '}{\cancel{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}} = \frac{{e}^{x} {\cos}^{2} y - 2 {e}^{y} \sin x \cos x}{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}$

$y ' = \frac{{e}^{x} {\cos}^{2} y - 2 {e}^{y} \sin x \cos x}{{e}^{y} {\sin}^{2} x + 2 {e}^{x} \sin y \cos y}$

God bless....I hope the explanation is useful.