How do you find the derivative of sin^2(lnx)?

Nov 1, 2015

Recall that ${\sin}^{2} \left(\ln x\right) = {\left[\sin \left(\ln x\right)\right]}^{2}$ and use the chain rule twice.

Explanation:

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(\ln x\right)\right) = \frac{d}{\mathrm{dx}} \left({\left[\sin \left(\ln x\right)\right]}^{2}\right)$

$= 2 \sin \left(\ln x\right) \left[\frac{d}{\mathrm{dx}} \left(\sin \left(\ln x\right)\right)\right]$

$= 2 \sin \left(\ln x\right) \left[\cos \left(\ln x\right) \frac{d}{\mathrm{dx}} \left(\ln x\right)\right]$

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

$= \frac{2 \sin \left(\ln x\right) \cos \left(\ln x\right)}{x}$

Which we may prefer to write as:

$= \sin \frac{2 \ln x}{x}$

Or, perhaps as

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