# How do you find (dy)/(dx) given e^siny+y=x^2?

Aug 30, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{{e}^{\sin} y \cos y + 1}$

#### Explanation:

Differentiate both sides. Recall that the chain rule will be put into effect.

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

This gives us:

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

Reapplying the chain rule:

${e}^{\sin} y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{\sin} y \cos y + 1\right) = 2 x$

Thus:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{{e}^{\sin} y \cos y + 1}$