# How do you implicitly differentiate 2=e^ysinx-x^2y^3 ?

Feb 29, 2016

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{3} - {e}^{y} \cos \left(x\right)}{{e}^{y} \sin \left(x\right) - 3 {x}^{2} {y}^{2}}$

#### Explanation:

You need to consider $y$ as a function of $x$ and so differentiate it accordingly. For example:
if you have ${y}^{2}$ differentiating you'll get $2 y \frac{\mathrm{dy}}{\mathrm{dx}}$ to consider the fact that $y$ represents a function of something!
in our case:
$0 = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(x\right) + {e}^{y} \cos \left(x\right) - 2 x {y}^{3} - {x}^{2} 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$
Here I used the Product Rule and every time I differentiated $y$ I needed to include $\frac{\mathrm{dy}}{\mathrm{dx}}$:

collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[{e}^{y} \sin \left(x\right) - {x}^{2} 3 {y}^{2}\right] = 2 x {y}^{3} - {e}^{y} \cos \left(x\right)$
and:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{3} - {e}^{y} \cos \left(x\right)}{{e}^{y} \sin \left(x\right) - {x}^{2} 3 {y}^{2}}$