Question

How do you differentiate ${{\left( \ln x \right)}^{\tan x}}$?

Answer 1

The given function is of the form ${{\left( f(x) \right)}^{g(x)}}$. We can’t differentiate these types of functions directly. We will need to make some changes to the function before differentiating it. We should know the derivative of the function $\ln x$ is $\dfrac{1}{x}$. Also, the derivative of the function $\tan x$ is ${{\sec }^{2}}x$.

Complete step by step answer:
Let $y={{\left( \ln x \right)}^{\tan x}}$. Taking log on both sides of the function, we get $\ln y=\ln \left( {{\left( \ln x \right)}^{\tan x}} \right)$. We can simplify this expression as $y=\tan x\ln \left( \ln x \right)$.
Differentiating both sides of this function, we get
$\dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left( \tan x\ln \left( \ln x \right) \right)}{dx}$
We know that the derivative of $\ln x$ is $\dfrac{1}{x}$, hence
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d\left( \tan x\ln \left( \ln x \right) \right)}{dx}$
Using product rule in the above expression, we can evaluate it as
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d\left( \tan x \right)}{dx}\ln \left( \ln x \right)+\tan x\dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}$
The derivative of $\tan x$ is ${{\sec }^{2}}x$, substituting it in the above equation, we get
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}={{\sec }^{2}}x\ln \left( \ln x \right)+\tan x\dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}$
As $\ln \left( \ln x \right)$ is a composite function of the form $f\left( g(x) \right)$. The composite functions are functions of the form $f\left( g(x) \right)$, their derivative is evaluated as, $\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}$.
Thus, $\dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}=\dfrac{d\left( \ln \left( \ln x \right) \right)}{d\left( \ln x \right)}\dfrac{d\left( \ln x \right)}{dx}$. Substituting the values of derivative, we get

& \Rightarrow \dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}=\dfrac{d\left( \ln \left( \ln x \right) \right)}{d\left( \ln x \right)}\dfrac{d\left( \ln x \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}=\dfrac{1}{\ln x}\dfrac{1}{x}=\dfrac{1}{x\ln x} \\\ \end{aligned}$$ substituting this expression in the derivative of the required function, we get $$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}={{\sec }^{2}}x\ln \left( \ln x \right)+\tan x\dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}$$ $$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}={{\sec }^{2}}x\ln \left( \ln x \right)+\tan x\dfrac{1}{x\ln x}$$ Multiplying y on both sides, we get $$\Rightarrow \dfrac{dy}{dx}=\left( {{\sec }^{2}}x\ln \left( \ln x \right)+\dfrac{\tan x}{x\ln x} \right)y$$ substituting the value of y in above expression, we get $$\Rightarrow \dfrac{dy}{dx}=\left( {{\sec }^{2}}x\ln \left( \ln x \right)+\dfrac{\tan x}{x\ln x} \right)\tan x\ln \left( \ln x \right)$$ **Note:** This is the standard procedure to evaluate the functions of the form $${{\left( f(x) \right)}^{g(x)}}$$. The steps are as follows, Step 1: Take a log on both sides of equation $$y={{\left( f(x) \right)}^{g(x)}}$$. Simplify the equation as $$\ln y=g(x)\ln \left( f(x) \right)$$. Step 2: Differentiate both sides as $$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d\left( g(x)\ln \left( f(x) \right) \right)}{dx}$$ Step 3: simplify the right-hand side using the product rule, and composite function derivative Step 4: multiply both sides by y, simplify the expression.