# How do you implicitly differentiate 9=-y/x+xsiny?

##### 1 Answer
Apr 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} \sin y + y}{x - {x}^{3} \cos y}$

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

$\text{differentiate " y/x" using the "color(blue)"quotient rule}$

$\text{differentiate " xsiny" using the "color(blue)"product rule}$

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

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

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

$\Rightarrow x \frac{\mathrm{dy}}{\mathrm{dx}} - y = {x}^{3} \cos y . \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{2} \sin y$

$\Rightarrow x \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{3} \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \sin y + y$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} \sin y + y}{x - {x}^{3} \cos y}$