Question

Differentiate ${\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\dfrac{x}{2}$ ?

Answer 1

Hint : In this question, first we have to make an identity of $\tan \left( {A - B} \right)$ with the trigonometric functions written in tan inverse. Then we have to differentiate it with respect to $\dfrac{x}{2}$.
Which we can do by differentiating both ${\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ and $\dfrac{x}{2}$ with respect to x.

Complete step-by-step answer :
In the given question, we have ${\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$.
Now,
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$
Now divide with $\cos x$ in both numerator and denominator.
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sin x}}{{\cos x}} - 1}}{{\dfrac{{\sin x}}{{\cos x}} + 1}}} \right)$
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\tan x - 1}}{{\tan x + 1}}} \right)$
As we know that $\tan \dfrac{\pi }{4} = 1$
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\tan x - \tan \dfrac{\pi }{4}}}{{1 + \tan x\tan \dfrac{\pi }{4}}}} \right)$
There is an identity in trigonometry as $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$.
On applying it,
$\Rightarrow {\tan ^{ - 1}}\left( {\tan \left( {x - \dfrac{\pi }{4}} \right)} \right)$
Now taking inverse of given function.
$\Rightarrow x - \dfrac{\pi }{4}$
Therefore, we can say that $x - \dfrac{\pi }{4} = {\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$
Now differentiate the above value with respect to x.
$\Rightarrow \dfrac{{d\left( {x - \dfrac{\pi }{4}} \right)}}{{dx}} = \dfrac{{dx}}{{dx}} - \dfrac{{d\left( {\dfrac{\pi }{4}} \right)}}{{dx}} = 1 - 0 = 1$
$\Rightarrow \dfrac{{d\left( {x - \dfrac{\pi }{4}} \right)}}{{dx}} = 1 - - - - \left( 1 \right)$
We know that differentiation of a constant value with respect to x is zero.
Now differentiate $\dfrac{x}{2}$ with respect to x.
$\Rightarrow \dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}} = \dfrac{1}{2} - - - - \left( 2 \right)$
Now divide equation $\left( 1 \right)\,and\,\left( 2 \right)$
$\Rightarrow \dfrac{{\dfrac{{d\left( {x - \dfrac{\pi }{4}} \right)}}{{dx}}}}{{\dfrac{{d\left( {\dfrac{x}{2}} \right)}}{{dx}}}} = \dfrac{1}{{\dfrac{1}{2}}}$
$\Rightarrow \dfrac{{d\left( {x - \dfrac{\pi }{4}} \right)}}{{d\left( {\dfrac{x}{2}} \right)}} = 2$
Now substitute the value $x - \dfrac{\pi }{4} = {\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$
$\Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)} \right)}}{{d\left( {\dfrac{x}{2}} \right)}} = 2$
Therefore, the differentiation of ${\tan ^{ - 1}}\left( {\dfrac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\dfrac{x}{2}$ is $2$ .
So, the correct answer is “2”.

Note : Differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables. The most common example is the rate change of displacement with respect to time, called velocity. The opposite of finding a derivative is anti-differentiation.
If x is a variable and y is another variable, then the rate of change of x with respect to y is given by $\dfrac{{dy}}{{dx}}$. This is the general expression of derivative of a function and is represented as $f'\left( x \right){\text{ }} = {\text{ }}\dfrac{{dy}}{{dx}}$, where $y{\text{ }} = {\text{ }}f\left( x \right)$ is any function.