How do you implicitly differentiate xy- 3yln(x-y)= 4/x?

Apr 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{4}{x} ^ 2 - y + \frac{3 y}{x - y}}{x - 3 \ln \left(x - y\right) + \frac{3 y}{x - y}}$

Explanation:

Differentiating both sides of

$x y - 3 y \ln \left(x - y\right) = \frac{4}{x}$

with respect to $x$ gives

$y + x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 \frac{\mathrm{dy}}{\mathrm{dx}} \ln \left(x - y\right) - \frac{3 y}{x - y} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \frac{4}{x} ^ 2$

so

$\left[x - 3 \ln \left(x - y\right) + \frac{3 y}{x - y}\right] \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4}{x} ^ 2 - y + \frac{3 y}{x - y}$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{4}{x} ^ 2 - y + \frac{3 y}{x - y}}{x - 3 \ln \left(x - y\right) + \frac{3 y}{x - y}}$