# How do you differentiate 3xy^2+cosy^2=2x^3+5?

Jan 12, 2017

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(3 x {y}^{2} + \cos {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(2 {x}^{3} + 5\right)$

$\frac{d}{\mathrm{dx}} 3 x {y}^{2} + \frac{d}{\mathrm{dx}} \cos {y}^{2} = \frac{d}{\mathrm{dx}} 2 {x}^{3} + \frac{d}{\mathrm{dx}} 5$

$\left[\frac{d}{\mathrm{dx}} 3 x {y}^{2}\right] + \frac{d}{\mathrm{dx}} \cos {y}^{2} = \frac{d}{\mathrm{dx}} 2 {x}^{3} + \frac{d}{\mathrm{dx}} 5$

$\left[3 x 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} 3\right] + \left(- 2 y \sin {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} + 0$
differentiate of $\sin {y}^{2}$ and ${\sin}^{2} y$ is different answer

$6 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} - \left(2 y \sin {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(6 x y - 2 y \sin {y}^{2}\right) = 6 {x}^{2} - 3 {y}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{2} - 3 {y}^{2}}{6 x y - 2 y \sin {y}^{2}} = \frac{3 \left(2 {x}^{2} - {y}^{2}\right)}{2 y \left(3 x - \sin {y}^{2}\right)}$