# How do you find the derivative of  (ln x) ^ cos x?

Jul 13, 2016

$y ' = {\left(\ln x\right)}^{\cos x} \left[- \sin x \cdot \ln \left(\ln x\right) + \frac{\cos x}{x \ln x}\right]$

#### Explanation:

For this particular, we'd have to use logarithmic differentiation, which works as follows:

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

Taking the natural log ($\ln$) of both sides yields

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

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

Since the next step is to take derivatives, the rules we're going to use is

$\frac{d}{\mathrm{dx}} \left[\ln u\right] = \frac{u '}{u}$

Differentiating both sides gives

$\frac{y '}{y} = - \sin x \cdot \ln \left(\ln x\right) + \cos x \cdot \frac{\left(\frac{1}{x}\right) \cdot \left(\frac{1}{\ln} x\right)}{\left(\frac{\cancel{\ln x}}{1}\right) \cdot \left(\frac{1}{\cancel{\ln x}}\right)}$

$\frac{y '}{y} = - \sin x \cdot \ln \left(\ln x\right) + \frac{\cos x}{x \ln x}$

But since we're looking for $y '$, we simply move $y$ to the right side of our equation to get

$y ' = y \left[- \sin x \cdot \ln \left(\ln x\right) + \frac{\cos x}{x \ln x}\right]$

Substituting $y$ back into the equation yields

$y ' = {\left(\ln x\right)}^{\cos x} \left[- \sin x \cdot \ln \left(\ln x\right) + \frac{\cos x}{x \ln x}\right]$