Question

What is the derivative of $\ln \left( {{\tan }^{2}}x \right)$?

Answer 1

In this method we need to find the derivative of the given function. For this we are going to use the substitution method and use the substitution $u={{\tan }^{2}}x$. First, we will differentiate the substitution $u={{\tan }^{2}}x$ with respect to $x$ by using the differentiation formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$. After using all the above formulas, we will get the value of $\dfrac{du}{dx}$. Now we will consider the given function and use the assumed substitution value and differentiate the equation. Here we will use the calculated value of $\dfrac{du}{dx}$ and differentiation formula $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ and simplify the equation to get the required result.

Complete step by step solution:
Given function is $\ln \left( {{\tan }^{2}}x \right)$.
Let us assume the substitution $u={{\tan }^{2}}x$. Substituting this value in the given function, then we will get
$\ln \left( {{\tan }^{2}}x \right)=\ln \left( u \right).....\left( \text{i} \right)$
Considering the substitution $u={{\tan }^{2}}x$ and differentiating the equation with respect to $x$, then we will have
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( {{\tan }^{2}}x \right)$
We have the differentiation formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$. Applying all the above formulas and simplifying the equation, then we will get
$\begin{aligned} & \dfrac{du}{dx}=2\tan x\times \dfrac{d}{dx}\left( \tan x \right) \\\ & \Rightarrow \dfrac{du}{dx}=2\tan x{{\sec }^{2}}x \\\ \end{aligned}$
Now considering the equation $\left( \text{i} \right)$ and differentiating the equation with respect to $x$, then we will have
$\dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=\dfrac{d}{dx}\left( \ln u \right)$
Applying the differentiation formulas $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ in the above equation, then we will get
$\dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=\dfrac{1}{u}\times \dfrac{du}{dx}$
Substituting the values $\dfrac{du}{dx}=2\tan x{{\sec }^{2}}x$, $u={{\tan }^{2}}x$ in the above equation, then we will have
$\dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=\dfrac{1}{{{\tan }^{2}}x}\times 2\tan x{{\sec }^{2}}x$
Cancelling the term $\tan x$ which is in both numerator and denominator, then we will get
$\dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=\dfrac{2{{\sec }^{2}}x}{\tan x}$

Hence the derivative of the given function $\ln \left( {{\tan }^{2}}x \right)$ is $\dfrac{2{{\sec }^{2}}x}{\tan x}$.

Note: We can also simplify the above result by using the trigonometric identity ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. From this identity we can write ${{\sec }^{2}}x=1+{{\tan }^{2}}x$, substituting this value in the obtained result then we will get
$\dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=2\dfrac{\left( {{\tan }^{2}}x+1 \right)}{\tan x}$
Simplifying the above equation by using some basic mathematical operations and the value $\cot x=\dfrac{1}{\tan x}$, then we will have
$\begin{aligned} & \dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=2\left[ \dfrac{{{\tan }^{2}}x}{\tan x}+\dfrac{1}{\tan x} \right] \\\ & \Rightarrow \dfrac{d}{dx}\left( \ln \left( {{\tan }^{2}}x \right) \right)=2\left( \tan x+\cot x \right) \\\ \end{aligned}$
So, we can also write the derivative of the given equation $\ln \left( {{\tan }^{2}}x \right)$ as $2\left( \tan x+\cot x \right)$