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

Dec 23, 2015

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

#### Explanation:

Use the chain rule.

First, treat ${\sin}^{2} \left(\ln x\right)$ as ${u}^{2}$ where $u = \sin \left(\ln x\right)$.

$\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = 2 u \cdot u '$

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

To find $\frac{d}{\mathrm{dx}} \left[\sin \left(\ln x\right)\right]$, treat it as $\sin \left(v\right)$ where $v = \ln x$.

$\frac{d}{\mathrm{dx}} \left[\sin \left(v\right)\right] = \cos \left(v\right) \cdot v '$

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

Also, $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$.

Plug this in:

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

Plug this in:

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