# How do you find the derivative of  e^(xy)=x/y?

Jan 6, 2017

Use the natural log on both side, separate the division into the difference of two logs, use the product rule on the left and the chain rule on the y terms.

#### Explanation:

Use the natural logarithm on both sides:

$\ln \left({e}^{x y}\right) = \ln \left(\frac{x}{y}\right)$

The inverse functions on the left cancel each other:

$x y = \ln \left(\frac{x}{y}\right)$

Division in the argument of a logarithm is the same as the difference between the dividend and the divisor:

$x y = \ln \left(x\right) - \ln \left(y\right)$

$x y + \ln \left(y\right) = \ln \left(x\right)$

Differentiate each term:

$\frac{d \left(x y\right)}{\mathrm{dx}} + \frac{d \left(\ln \left(y\right)\right)}{\mathrm{dx}} = \frac{d \left(\ln \left(x\right)\right)}{\mathrm{dx}}$

Use the product rule combined with the chain rule on the first term:

$y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d \left(\ln \left(y\right)\right)}{\mathrm{dx}} = \frac{d \left(\ln \left(x\right)\right)}{\mathrm{dx}}$

Use the chain rule on the second term:

$y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(\ln \left(x\right)\right)}{\mathrm{dx}}$

The last term is trivial:

$y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x}$

Subtract y from both sides:

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Make common denominators on both sides:

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

Multiply both sides by $\frac{y}{x y + 1}$:

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

Use the distributive property on the numerator and denominator:

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