How do you use the chain rule to differentiate \[\ln (\tan x)\]?

Hint: The given expression \[\ln (\tan x)\] is to be differentiated using chain rule. It implies that we will carry out the differentiation in a stepwise fashion. We will first differentiate the logarithm function without differentiating the angle (tangent function) and then we will multiply it with the differentiation of the tangent function. Hence, we have the differentiation of the given expression.
For eg – if we want to differentiate sin x with respect to x, we get,
\[\sin x=\dfrac{d}{dx}(\sin x)=\cos x\]
Chain rule refers to the sequence in which the expression will be differentiated. Let’s understand it using an example, it is shown as follows:
\[f(g(x))=f'(g(x)).g'(x)\]
Here, \[g(x)\] can be thought of as a composite function
So, first the function ‘f’ is differentiated and then its composite function \[g(x)\] is differentiated.
For example – to find the derivative of \[\sin ({{x}^{3}})\]
Let, \[f(g(x))=\sin ({{x}^{3}})\]
So, \[f(x)=\sin x\] and \[g(x)={{x}^{3}}\]
Differentiating we get, \[f'(x)=\cos x\] and \[g'(x)=3{{x}^{2}}\], as we know \[\dfrac{d}{dx}(\sin x)=\cos x\] and \[\dfrac{d}{dx}({{x}^{3}})=3{{x}^{2}}\] respectively.
Applying chain rule we have,
\[f(g(x))=f'(g(x)).g'(x)\]
Substituting in it we have,
\[\Rightarrow \dfrac{d}{dx}(\sin ({{x}^{3}}))=\cos ({{x}^{3}}).3{{x}^{2}}\]

Complete step by step solution:
According to the question we have to differentiate \[\ln (\tan x)\] using chain rule.
So, let’s say, \[f(g(x))=\ln (\tan x)\]
So, \[f(x)=\ln x\] and \[g(x)=\tan x\]
Differentiating we get, \[f'(x)=\dfrac{1}{x}\] and \[g'(x)={{\sec }^{2}}x\], as we know \[\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\] and \[\dfrac{d}{dx}(\tan x)={{\sec }^{2}}x\] respectively.
Applying chain rule we have,
\[f(g(x))=f'(g(x)).g'(x)\]
Substituting in it we have,
\[\Rightarrow \dfrac{d}{dx}(\ln (\tan x))=\dfrac{1}{\tan x}.{{\sec }^{2}}x\]

Note: we can also carry out the differentiation like-
Let \[h(x)=\ln (\tan x)\]
\[\dfrac{d}{dx}(h(x))=\dfrac{d}{dx}(\ln (\tan x))\]
\[=\dfrac{d}{d\tan x}(\ln (\tan x))\times \dfrac{d\tan x}{dx}(\tan x)\]
\[=\dfrac{1}{\tan x}.{{\sec }^{2}}x\]
As we know, \[\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\] and \[\dfrac{d}{dx}(\tan x)={{\sec }^{2}}x\]
Therefore, \[\dfrac{d}{dx}(\ln (\tan x))=\dfrac{1}{\tan x}.{{\sec }^{2}}x\] .