# How do you find the derivative of ln x / ln y = ln(x-y)?

Jun 17, 2016

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

#### Explanation:

As $\frac{\ln x}{\ln y} = \ln \left(x - y\right)$, using chain rule and differentiating both sides

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

(note that we have used quotient rule on LHS)

$\left(\ln \frac{y}{x} - \ln \frac{x}{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}\right) = {\left(\ln y\right)}^{2} / \left(x - y\right) - \frac{\mathrm{dy}}{\mathrm{dx}} \left({\left(\ln y\right)}^{2} / \left(x - y\right)\right)$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left({\left(\ln y\right)}^{2} / \left(x - y\right)\right) - \ln \frac{x}{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln y\right)}^{2} / \left(x - y\right) - \ln \frac{y}{x}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left({\left(\ln y\right)}^{2} / \left(x - y\right) - \ln \frac{x}{y}\right) = {\left(\ln y\right)}^{2} / \left(x - y\right) - \ln \frac{y}{x}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(\ln y\right)}^{2} / \left(x - y\right) - \ln \frac{y}{x}}{{\left(\ln y\right)}^{2} / \left(x - y\right) - \ln \frac{x}{y}}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {\left(\ln y\right)}^{2} - \left(x - y\right) \ln y}{x {\left(\ln y\right)}^{2} - \left(x - y\right) \ln x} \times \frac{y}{x}$