Question

How do you find the derivative of ${{x}^{\tan x}}$?

Answer 1

First we will assume the given equation to a variable. For the given equation we will apply logarithmic function in order to convert the given equation in form of the multiplication. Now we will apply differentiation to the obtained equation. Here we will use several formulas which are $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$, $\dfrac{d}{dx}\left( uv \right)=u{{v}^{'}}+v{{u}^{'}}$. By using the above formulas, we will get the required result.

Complete step by step answer:
Given that, ${{x}^{\tan x}}$.
Let $y={{x}^{\tan x}}$
Applying logarithmic function to the both sides of the above equation, then we will get
$\log \left( y \right)=\log \left( {{x}^{\tan x}} \right)$
We have logarithmic formula $\log \left( {{a}^{b}} \right)=b\log \left( a \right)$, then we will get
$\log y=\tan x.\log x$
Differentiating the above equation with respect to the $x$, then we will have
$\dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left( \tan x.\log x \right)$
Applying the $uv$ rule or $\dfrac{d}{dx}\left( uv \right)=u{{v}^{'}}+v{{u}^{'}}$ in the above equation, then we will get
$\Rightarrow \dfrac{d}{dx}\left( \log y \right)=\tan x\left[ \dfrac{d}{dx}\left( \log x \right) \right]+\log x\left[ \dfrac{d}{dx}\left( \tan x \right) \right]$
Using the formulas $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$, $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$ in the above equation, then we will get
$\begin{aligned} & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\tan x\left[ \dfrac{1}{x} \right]+\log x\left[ {{\sec }^{2}}x \right] \\\ & \Rightarrow \dfrac{dy}{dx}=y\left[ {{\sec }^{2}}x\log x+\dfrac{\tan x}{x} \right] \\\ \end{aligned}$
We have assumed that $y={{x}^{\tan x}}$, so substituting this value in the above equation, then we will get
$\Rightarrow \dfrac{dy}{dx}={{x}^{\tan x}}\left( {{\sec }^{2}}x\log x+\dfrac{\tan x}{x} \right)$

$\therefore$The derivative of the equation ${{x}^{\tan x}}$ is ${{x}^{\tan x}}\left( {{\sec }^{2}}x\log x+\dfrac{\tan x}{x} \right)$.

Note: In the problem we have the function $\tan x$ as the power of the $x$. So, we have applied logarithmic function and thereafter we have used the appropriate formulas. But when the function $\tan x$ is in multiplication of $x$, then there is no need of applying a logarithmic function just use the $uv$ rule and simplify the obtained equation to get the result.