# How do you use logarithmic differentiation to find the derivative of y=(tanx)^(1/x)?

Mar 27, 2015

$y = {\left(\tan x\right)}^{\frac{1}{x}}$

$\ln y = \ln \left({\left(\tan x\right)}^{\frac{1}{x}}\right)$

$\ln y = \frac{1}{x} \ln \left(\tan x\right)$ Differentiate implicitly.

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

At this point it's nice to simplify using $\frac{1}{\tan} x = \cot x$ and $\cot x \cdot \sec x = \csc x$

So,
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ 2 \ln \left(\tan \left(x\right)\right) + \frac{1}{x} \left(\csc x \sec x\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\tan x\right)}^{\frac{1}{x}} \left(- \frac{1}{x} ^ 2 \ln \left(\tan \left(x\right)\right) + \frac{1}{x} \left(\csc x \sec x\right)\right)$ $\text{ }$ (Not pretty, but correct.)

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\tan x\right)}^{\frac{1}{x}} \left(x \left(\csc x \sec x\right) - \ln \left(\tan \left(x\right)\right)\right) \cdot \frac{1}{x} ^ 2$ $\text{ }$ (Isn't a whole lot better.)