# How do you differentiate e^(x^(2)y) = 2x + 2y?

May 24, 2015

Take one side at a time.

Let $f \left(x , y\right) = {e}^{{x}^{2} y}$.

${\left[\frac{\mathrm{df}}{\mathrm{dx}}\right]}_{\text{left}} = \left({e}^{{x}^{2} y}\right) \left[{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot 2 x\right]$
(chain rule, product rule)

${\left[\frac{\mathrm{df}}{\mathrm{dx}}\right]}_{\text{right}} = 2 + 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
(chain rule with $\frac{\mathrm{dy}}{\mathrm{dx}}$)

Now, set them equal...
$\left({e}^{{x}^{2} y}\right) \left[{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot 2 x\right] = 2 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

...Multiply through.
${x}^{2} {e}^{{x}^{2} y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y {e}^{{x}^{2} y} = 2 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

You can now isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$. Move everything related to each other to each side.

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

Factor out common terms:
$2 x y {e}^{{x}^{2} y} - 2 = \frac{\mathrm{dy}}{\mathrm{dx}} \left[2 - {x}^{2} {e}^{{x}^{2} y}\right]$

Divide bracketed stuff over:
$\frac{2 x y {e}^{{x}^{2} y} - 2}{2 - {x}^{2} {e}^{{x}^{2} y}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

...And simplify so it looks nicer.
$- \frac{2 - 2 x y {e}^{{x}^{2} y}}{2 - {x}^{2} {e}^{{x}^{2} y}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Wolfram Alpha lists the answer as:
$\frac{2 - 2 x y {e}^{{x}^{2} y}}{{x}^{2} {e}^{{x}^{2} y} - 2} = \frac{\mathrm{dy}}{\mathrm{dx}}$