# How do you differentiate e^sqrt(xy)-sqrt(xy)=8?

Dec 20, 2015

See explanation.

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

where $c$ is the square of the positive root of ${e}^{t} - t - 8 = 0$

#### Explanation:

Let $t = \sqrt{x y}$

Then ${e}^{t} - t = 8$

This has two roots ${t}_{1} \approx 2.33559$, ${t}_{2} \approx - 7.999664$

Numerical approximations for ${t}_{1}$ and ${t}_{2}$ can be found using Newton's method or similar.

We can discard ${t}_{2}$ since $t = \sqrt{x y}$ is the non-negative square root.

Let $c = {t}_{1}^{2}$

Then the original equation simplifies to:

$x y = c$

So $y = \frac{c}{x}$ and $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{c}{{x}^{2}}$