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

$y ' = \frac{2 x {y}^{3}}{{e}^{y} \cdot \cos y + {e}^{y} \cdot \sin y - 3 {x}^{2} {y}^{2}}$

#### Explanation:

From the given equation

$2 = {e}^{y} \sin y - {x}^{2} {y}^{3}$

Differentiate both sides of the equation with respect to $x$

$\frac{d}{\mathrm{dx}} \left(2\right) = \frac{d}{\mathrm{dx}} \left({e}^{y} \sin y - {x}^{2} {y}^{3}\right)$

$0 = {e}^{y} \cdot \frac{d}{\mathrm{dx}} \left(\sin y\right) + \sin y \cdot \frac{d}{\mathrm{dx}} \left({e}^{y}\right) - \left[{x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({y}^{3}\right) + {y}^{3} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right]$

$0 = {e}^{y} \cdot \left(\cos y\right) y ' + \sin y \cdot \left({e}^{y}\right) \cdot y ' - \left[{x}^{2} \left(3 {y}^{2}\right) y ' + \left({y}^{3}\right) \left(2 x\right)\right]$

$0 = {e}^{y} \cdot \left(\cos y\right) y ' + \sin y \cdot \left({e}^{y}\right) \cdot y ' - {x}^{2} \left(3 {y}^{2}\right) y ' - \left({y}^{3}\right) \left(2 x\right)$

$0 = \left[{e}^{y} \cdot \left(\cos y\right) + \sin y \cdot \left({e}^{y}\right) - {x}^{2} \left(3 {y}^{2}\right)\right] y ' - \left({y}^{3}\right) \left(2 x\right)$

$\left({y}^{3}\right) \left(2 x\right) = \left[{e}^{y} \cdot \left(\cos y\right) + \sin y \cdot \left({e}^{y}\right) - {x}^{2} \left(3 {y}^{2}\right)\right] y '$

$y ' = \frac{2 x {y}^{3}}{{e}^{y} \cdot \cos y + {e}^{y} \cdot \sin y - 3 {x}^{2} {y}^{2}}$

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