# How do you use implicit differentiation to find dy/dx given xsiny+x^2cosy=1?

##### 1 Answer
Oct 31, 2016

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

#### Explanation:

Differentiation of the expression is determined by using the product rule differentiation

$\textcolor{b r o w n}{\frac{d}{\mathrm{dx}} \left(u \left(x\right) + v \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(u \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(v \left(x\right)\right)}$

color(red)(d/dx(u(x)xxv(x))
color(red)(=(du(x))/dxxxv(x)+(dv(x))/dxxxu(x)

Applying the differentiation of trigonometric functions
$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x}$
$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x}$

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

$\Rightarrow \textcolor{b r o w n}{\frac{d}{\mathrm{dx}} \left(x \sin y\right) + \frac{d}{\mathrm{dx}} \left({x}^{2} \cos y\right) = 0}$

rArrcolor(red)(((dx)/dxxxsiny+(dsiny)/dxxxx)+((dx^2)/dxxxcosy+(dcosy)/dxxxx^2)=0

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

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

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

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