# How do you differentiate f(x)=e^(cos(lnx)) using the chain rule?

Dec 7, 2015

The answer is $f ' \left(x\right) = {e}^{\cos \left(\ln \left(x\right)\right)} \cdot - \sin \left(\ln \left(x\right)\right) \cdot \frac{1}{x} = - \frac{\sin \left(\ln \left(x\right)\right) {e}^{\cos \left(\ln \left(x\right)\right)}}{x}$

#### Explanation:

Here are the details, based on using the formula $\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$ twice:

$f ' \left(x\right) = {e}^{\cos \left(\ln \left(x\right)\right)} \cdot \frac{d}{\mathrm{dx}} \left(\cos \left(\ln \left(x\right)\right)\right)$

$= {e}^{\cos \left(\ln \left(x\right)\right)} \cdot - \sin \left(\ln \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right)$

$= {e}^{\cos \left(\ln \left(x\right)\right)} \cdot - \sin \left(\ln \left(x\right)\right) \cdot \frac{1}{x} = - \frac{\sin \left(\ln \left(x\right)\right) {e}^{\cos \left(\ln \left(x\right)\right)}}{x}$