# Question #f1456

Jan 28, 2018

$\cot \frac{x}{\ln} \left(10\right)$

#### Explanation:

Remember the change of base formula: ${\log}_{a} b = \frac{\ln \left(b\right)}{\ln \left(a\right)}$.

Here, $a = 10 ,$ and $b = \sin \left(x\right)$.

So input: $\frac{d}{\mathrm{dx}} \left(\frac{\ln \left(\sin \left(x\right)\right)}{\ln \left(10\right)}\right)$

Take the constant out. Basically, the part that does not affect the variable $x$. For example, while solving for $\frac{d}{\mathrm{dx}} 2 {x}^{2}$, you could take the $2$ out as it would not affect the final value.

$\frac{1}{\ln \left(10\right)} \frac{d}{\mathrm{dx}} \ln \left(\sin \left(x\right)\right)$.

Now apply the chain rule. $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Here, $f = \ln u$, and $u = \sin \left(x\right)$.

$\frac{d}{\mathrm{du}} \ln u = \frac{1}{u}$

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

So $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{1}{u} \cdot \cos \left(x\right) = \cos \frac{x}{u}$.

But remember, $u = \sin \left(x\right)$

So it becomes $\cos \frac{x}{\sin} \left(x\right) = \cot \left(x\right)$

Multiply this by the constant, $\frac{1}{\ln} \left(10\right) \cdot \cot \left(x\right) = \cot \frac{x}{\ln} \left(10\right)$

The derivative of $\log \left(\sin \left(x\right)\right)$ is $\cot \frac{x}{\ln} \left(10\right)$