# How do you find dy/dx for the function; cos(x^2+2Y)+xe^Y^2=1?

Apr 11, 2017

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

#### Explanation:

Using the implicit function theorem:

$f \left(x , y\right) = \cos \left({x}^{2} + 2 y\right) + x {e}^{{y}^{2}} = 1$

$\mathrm{df} = 0 = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$

So:

${f}_{x} = - 2 x \sin \left({x}^{2} + 2 y\right) + {e}^{{y}^{2}}$

${f}_{y} = - 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}$

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