Question

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

Answer 1

Hint: 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
\[\begin{align}
  & \Rightarrow \dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}=\dfrac{d\left( \ln \left( \ln x \right) \right)}{d\left( \ln x \right)}\dfrac{d\left( \ln x \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( \ln \left( \ln x \right) \right)}{dx}=\dfrac{1}{\ln x}\dfrac{1}{x}=\dfrac{1}{x\ln x} \\
\end{align}\]
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.