# What is the derivative of sinx^tanx?

Mar 10, 2018

$\cos \left({x}^{\tan \left(x\right)}\right) {x}^{\tan \left(x\right)} \left(\tan \frac{x}{x} + \ln \left(x\right) \cdot {\sec}^{2} \left(x\right)\right)$
[Assuming you meant $\sin \left({x}^{\tan \left(x\right)}\right)$]

#### Explanation:

We have:

$\frac{d}{\mathrm{dx}} \left[\sin \left({x}^{\tan \left(x\right)}\right)\right]$ We use the chain rule:

$\frac{d}{\mathrm{dx}} \left[g \left(h \left(x\right)\right)\right] = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

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

$\implies \cos \left({x}^{\tan \left(x\right)}\right) \cdot \frac{d}{\mathrm{dx}} \left[{x}^{\tan \left(x\right)}\right]$

Here is a rule that you may not be familiar with:

$\frac{d}{\mathrm{dx}} \left[{\left(f \left(x\right)\right)}^{g \left(x\right)}\right] = {\left(f \left(x\right)\right)}^{g \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(f \left(x\right)\right) \cdot g \left(x\right)\right]$

$\implies \cos \left({x}^{\tan \left(x\right)}\right) \cdot {x}^{\tan \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(x\right) \cdot \tan \left(x\right)\right]$

Use the product rule:

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

$\implies \cos \left({x}^{\tan \left(x\right)}\right) {x}^{\tan \left(x\right)} \left(\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] \cdot \tan \left(x\right) + \ln \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right)\right)$

Remember that:

$\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = {\sec}^{2} \left(x\right)$

$\implies \cos \left({x}^{\tan \left(x\right)}\right) {x}^{\tan \left(x\right)} \left(\frac{1}{x} \cdot \tan \left(x\right) + \ln \left(x\right) \cdot {\sec}^{2} \left(x\right)\right)$

$\implies \cos \left({x}^{\tan \left(x\right)}\right) {x}^{\tan \left(x\right)} \left(\tan \frac{x}{x} + \ln \left(x\right) \cdot {\sec}^{2} \left(x\right)\right)$

Mar 10, 2018

${\left(\sin x\right)}^{\tan x} \cdot \left(\frac{\left(\cos x\right) \left(\tan x\right)}{\sin x} + \ln \left(\sin x\right) \cdot {\sec}^{2} x\right)$
[Assuming you meant ${\left(\sin x\right)}^{\tan x}$]

#### Explanation:

We have:

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

We use a rule you may be unfamiliar with:

$\frac{d}{\mathrm{dx}} \left[{\left(f \left(x\right)\right)}^{g \left(x\right)}\right] = {\left(f \left(x\right)\right)}^{g \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(f \left(x\right)\right) \cdot g \left(x\right)\right]$

Therefore, we have:

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

The product rule:

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

$\implies {\left(\sin x\right)}^{\tan x} \cdot \left(\frac{d}{\mathrm{dx}} \left[\ln \left(\sin x\right)\right] \cdot \tan x + \ln \left(\sin x\right) \cdot \frac{d}{\mathrm{dx}} \left[\tan x\right]\right)$

Remember that:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$

$\implies {\left(\sin x\right)}^{\tan x} \cdot \left(\frac{1}{\sin x} \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right) \cdot \tan x + \ln \left(\sin x\right) \cdot {\sec}^{2} x\right)$

Another thing to remember here:

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

$\implies {\left(\sin x\right)}^{\tan x} \cdot \left(\frac{1}{\sin x} \cdot \cos x \cdot \tan x + \ln \left(\sin x\right) \cdot {\sec}^{2} x\right)$ Simplify.

$\implies {\left(\sin x\right)}^{\tan x} \cdot \left(\frac{\left(\cos x\right) \left(\tan x\right)}{\sin x} + \ln \left(\sin x\right) \cdot {\sec}^{2} x\right)$