# What is the implicit derivative of y-30=e^y-x^2+ye^x ?

Dec 30, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - y {e}^{x}}{{e}^{y} + {e}^{x} - 1}$

#### Explanation:

The only tricky part of implicit differentiation is remembering to use the chain rule with anything involving $y$.

color(red)y-color(blue)30=color(green)(e^y)-color(orange)(x^2)+color(maroon)(ye^x

color(red)(d/dx(y)=dy/dx

color(blue)(d/dx(30)=0

color(green)(d/dx(e^y)=e^y*dy/dx

color(orange)(d/dx(x^2)=2x

Product rule coming up:

color(maroon)(d/dx(ye^x)=dy/dx*e^x+y*d/dx(e^x)=dy/dx*e^x+ye^x

Combine the derivatives.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{y} - 2 x + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{x} + y {e}^{x}$

Isolate the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms and solve.

$2 x - y {e}^{x} = \frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{y} + {e}^{x} - 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - y {e}^{x}}{{e}^{y} + {e}^{x} - 1}$