Question

Find the derivative of $y = \ln \left( {\tan x} \right)$.

Answer 1

Hint:We know Chain Rule: $\left( {f(g(x))} \right) = f'(g(x))g'(x)$
By using Chain rule we can solve this problem. Since we cannot find the direct derivative of the given question we have to use the chain rule. So we must convert our question in the form of the equation above such that we have to find the values of every term in the above equation and substitute it back. In that way we would be able to find the solution for the given question.

Complete step by step solution:
Given
$y = \ln \left( {\tan x} \right)........................\left( i \right)$
So according to our question we need to find \[\dfrac{{dy}}{{dx}} = \dfrac{{d\ln \left( {\tan x}
\right)}}{{dx}}\]
Thus here we can use chain rule to find the derivative since we can’t find the derivative with any direct equation.
Now we know that chain rule is:\[f(g(x)) = f'(g(x))g'(x).......................\left( {ii} \right)\]
Such that on comparing (ii), if:
\[f\left( x \right) = \ln \left( x \right)\,\,{\text{and}}\,\,g\left( x \right) = \tan \left( x
\right).......................\left( {iii} \right)\]
Now to apply chain rule, let:
$u = \tan x$
Then we can say that $f\left( {g\left( x \right)} \right) = \ln u.......................\left( {iv} \right)$
Now on comparing to (ii) we can write:
\[
\dfrac{{dy}}{{dx}} = \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = f'(g(x))g'(x) \\
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\ln u}
\right)\dfrac{d}{{dx}}\left( {\tan x} \right)..............................\left( v \right) \\
\]
On observing (v) we can say that the derivative of $\ln u$with respect to $du$ would be $\dfrac{1}{u}$.
So on substituting we get:
\[
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\ln u}
\right)\dfrac{d}{{dx}}\left( {\tan x} \right) \\
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{1}{u}\dfrac{d}{{dx}}\left( {\tan x}
\right).................................\left( {vi} \right) \\
\]
Substituting $u = \tan x$back to (vi) we get:
\[ \Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{1}{{\tan x}}\dfrac{d}{{dx}}\left( {\tan x} \right).......................\left( {vii} \right)\]
Also we know that the derivative of $\tan x$ with respect to $dx$ would be ${\sec
^2}x$i.e.\[\dfrac{d}{{dx}}\tan x = {\sec ^2}x\]
Substituting this in (vii) we get:
\[ \Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{1}{{\tan x}}{\sec ^2}x..............\left(
{viii} \right)\]
Also we know that:
$
\tan x = \dfrac{{\sin x}}{{\cos x}}\;\; \Rightarrow \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}
\\
\sec x = \dfrac{1}{{\cos x}} \\
$
On substituting these values to (viii) we get:
\[
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{{\cos x}}{{\sin x}} \times \left(
{\dfrac{1}{{{{\cos }^2}x}}} \right) \\
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \dfrac{1}{{\sin x}} \times \left(
{\dfrac{1}{{\cos x}}} \right) \\
\Rightarrow \dfrac{{d\ln \left( {\tan x} \right)}}{{dx}} = \csc x \times \sec x..................\left( {ix}
\right) \\
\]
Therefore we can say that the derivative of$y = \ln \left( {\tan x} \right)$ is\[\csc x \times \sec x\].

Note:
Some of the important trigonometric formulas:
\[
\dfrac{d}{{dx}}\tan x = {\sec ^2}x \\
\dfrac{d}{{dx}}\ln x = \dfrac{1}{x} \\
\]
$
\tan x = \dfrac{{\sin x}}{{\cos x}} \\
\cos x = \dfrac{1}{{\sec x}} \\
\sin x = \dfrac{1}{{\csc x}} \\
\tan x = \dfrac{1}{{\cot x}} \\
$
Chain rule is mainly used for finding the derivative of a composite function. Also care must be taken while using chain rule since it should be applied only on composite functions and applying chain rule that isn’t composite may result in a wrong derivative. Also common logarithmic functions are log functions with base 10, and natural logarithmic functions are log functions with base ‘e’. Natural logarithmic functions can be represented by $\ln x\;{\text{or}}\;{\log _e}x$.