Feb 13, 2018

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

Explanation:

Express ${x}^{\tan} x$ as power of e:

${x}^{\tan} x = {e}^{\ln} \left({x}^{\tan} x\right) = {e}^{\ln x \tan x}$

$= \frac{d}{\mathrm{dx}} {e}^{\ln x \tan x}$

Using the chain rule, $\frac{d}{\mathrm{dx}} {e}^{\ln x \tan x} = \frac{{\mathrm{de}}^{u}}{\mathrm{du}} \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) ,$ where $u = \ln x \tan x$ and $\frac{d}{\mathrm{du}} \left({e}^{u}\right) = {e}^{u}$

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

Express ${e}^{\ln x \tan x}$ as a power of x:
${e}^{\ln x \tan x} = {e}^{\ln} \left({x}^{\tan} x\right) = {x}^{\tan} x$

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

Use the product rule, $\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
, where $u = \ln x$ and $v = \tan x$

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

The derivate of $\tan x$ is ${\sec}^{2} x$

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

The derivative of $\ln x$ is $\frac{1}{x}$

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

Feb 13, 2018

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

Explanation:

We shall use logarithmic differentiation - that is, we will take the natural log of both sides and differentiate implicitly w.r.t $x$

Given: $y = {x}^{\tan} \left(x\right)$

Take the natural log ($\ln$) of both sides:

$\ln \left(y\right) = \ln \left({x}^{\tan} \left(x\right)\right)$

Applying the power rule of natural log $\ln {\left(a\right)}^{b} = b \cdot \ln \left(a\right)$

$\ln \left(y\right) = \tan \left(x\right) \cdot \ln \left(x\right)$

Differentiate both sides implicitly w.r.t $x$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{b l u e}{{\sec}^{2} \left(x\right) \ln \left(x\right) + \tan \frac{x}{x}}$ (See work below)

To differentiate the RHS, we will need to use the product rule!

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

Let $f \left(x\right) = \tan \left(x\right)$ and $g \left(x\right) = \ln \left(x\right)$

Thus, $f ' \left(x\right) = {\sec}^{2} \left(x\right)$ and $g ' \left(x\right) = \frac{1}{x}$

By the product rule: $\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + f \left(x\right) g \left(x\right)$

Substituting we get:

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

Simplifying...

$\frac{d}{\mathrm{dx}} \left[\tan \left(x\right) \cdot \ln \left(x\right)\right] = {\sec}^{2} \left(x\right) \cdot \ln \left(x\right) + \tan \frac{x}{x}$

Going back to what we had before:

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(x\right) \ln \left(x\right) + \tan \frac{x}{x}$

We want to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ so we multiply both sides by $y$

$\cancel{\textcolor{red}{y}} \cdot \frac{1}{\cancel{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \left({\sec}^{2} \left(x\right) \ln \left(x\right) + \tan \frac{x}{x}\right) \cdot \textcolor{red}{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({\sec}^{2} \left(x\right) \ln \left(x\right) + \tan \frac{x}{x}\right) \cdot \textcolor{red}{y}$

We want to write everything in terms of $x$ but we have this $\textcolor{red}{y}$ in the way. You may recall that $\textcolor{red}{y}$ is given to us in the very beginning. $\textcolor{red}{y = {x}^{\tan} \left(x\right)}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left({\sec}^{2} \left(x\right) \ln \left(x\right) + \tan \frac{x}{x}\right) \cdot {x}^{\tan} \left(x\right)$