# How do you find (dy)/(dx) given x^2sinx+y^2cosy=1?

Nov 22, 2016

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

#### Explanation:

differentiate both sides with respect to $\text{x}$. For the terms on the $\text{ } L H S$ the product rule will have to be employed twice.

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

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

$2 x \sin x + {x}^{2} \cos x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} \cos y + {y}^{2} \left(- \sin y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

now rearrange for ""(dy)/(dx)

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

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