# What is the derivative of  (lnx)^(sinx)?

Jan 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln \left(x\right)\right)}^{\sin} \left(x\right) \left\{\sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)\right\}$

#### Explanation:

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

To find the derivative of such a problem we need to take logarithms on both the sides.

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

This step is done to move the exponent to the front of the equation.
as color(brown)(ln(a^n) = n*ln(a)

$\ln \left(y\right) = \sin \left(x\right) \cdot \ln \left(\ln \left(x\right)\right)$

Now let us differentiate both sides with respect to $x$

We shall use product rule color(blue)((uv)'=uv'+vu'

(1/y)dy/dx = sin(x)d/dx(ln(ln(x))) + ln(ln(x))d/dx(sin(x)

$\textcolor{b l u e}{\text{Derivative of ln(ln(x)) to be done using chain rule}}$

(1/y)dy/dx = sin(x)(1/ln(x))*d/dx(ln(x)+ln(ln(x))(cos(x))

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln \left(x\right)\right)}^{\sin} \left(x\right) \left\{\sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)\right\}$