# How do you implicitly differentiate  ysinx^2 = xsin(e^y)?

Apr 2, 2018

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

#### Explanation:

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

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

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

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