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

##### 1 Answer
Mar 27, 2016

Take the derivative with respect to $x$ on both sides an rearrange to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{y} \sin 2 x}{2 {\sin}^{2} y - {e}^{y} {\sin}^{2} x}$

#### Explanation:

Begin by taking the derivative with respect to $x$ of both sides:
$\frac{d}{\mathrm{dx}} \left(9\right) = \frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x - {\cos}^{2} y\right)$
On the left side, we have the derivative of a constant - which is $0$:
$0 = \frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x - {\cos}^{2} y\right)$

The sum rule says we can break the right side down to:
$\frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x\right) - \frac{d}{\mathrm{dx}} \left({\cos}^{2} y\right)$

First Term
We are trying to find $\frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x\right)$. This involves using the product rule, which says:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
In our case, $u = {e}^{y}$ and $v = {\sin}^{2} x$. Taking the derivatives of these two, we see:
$u = {e}^{y} \to u ' = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$
$v = {\sin}^{2} x = {\left(\sin x\right)}^{2} \to v ' = 2 \sin x \cos x = \sin 2 x$

Substituting into the product rule formula,
$\frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x\right) = \left({e}^{y}\right) ' \left({\sin}^{2} x\right) + \left({e}^{y}\right) \left({\sin}^{2} x\right) '$
$\textcolor{w h i t e}{X} = {e}^{y} {\sin}^{2} x \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} \sin 2 x$

Second Term
Now we have to find $\frac{d}{\mathrm{dx}} \left({\cos}^{2} y\right)$. Start by rewriting it as ${\left(\cos y\right)}^{2}$. Using the chain rule, we have to take the derivative of the inside term, $\cos y$ (which is $- \sin y \frac{\mathrm{dy}}{\mathrm{dx}}$), and multiply it by the whole derivative (i.e. the derivative of ${\left(\cos y\right)}^{2}$, which is $- 2 \sin y$). Doing so gives the result:
$- 2 \sin y \cdot - \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

Finally, we can combine our two results into one. Referencing the original equation,
$0 = \frac{d}{\mathrm{dx}} \left({e}^{y} {\sin}^{2} x\right) - \frac{d}{\mathrm{dx}} \left({\cos}^{2} y\right)$
Substituting what we found,
$0 = {e}^{y} {\sin}^{2} x \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} \sin 2 x - 2 {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

And doing a little algebra to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we find:
$0 = {e}^{y} {\sin}^{2} x \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} \sin 2 x - 2 {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$
$2 {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} {\sin}^{2} x \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \sin 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 {\sin}^{2} y - {e}^{y} {\sin}^{2} x\right) = {e}^{y} \sin 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{y} \sin 2 x}{2 {\sin}^{2} y - {e}^{y} {\sin}^{2} x}$