# What is the derivative of sin^2(lnx)?

Apr 3, 2015

$f \left(x\right) = {\sin}^{2} \left(\ln x\right) = {\left(\sin \left(\ln x\right)\right)}^{2}$

To find $f ' \left(x\right)$, use the chain rule, twice.

$f ' \left(x\right) = 2 \left(\sin \left(\ln x\right)\right) \cdot \left[\cos \left(\ln x\right) \cdot \frac{1}{x}\right]$

(The factor in the brackets is the derivative of $\sin \left(\ln x\right)$)

Clean up the answer however you want.

$f ' \left(x\right) = \frac{2}{x} \sin \left(\ln x\right) \cos \left(\ln x\right) \frac{1}{x} \sin \left(2 \ln x\right) = \frac{1}{x} \sin \left(\ln {x}^{2}\right)$