# How do you differentiate y=(lnx)^tanx?

Dec 9, 2016

Use logarithmic differentiation.

#### Explanation:

Use the natural logarithm on both sides:

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

Use the property of logarithms $\ln \left({a}^{b}\right) = \left(b\right) \ln \left(a\right)$

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

Differentiate:

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

Multyply both sides by y:

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

Substitute $\ln {\left(x\right)}^{\tan} \left(x\right)$ for y:

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