Question

If $y = {\tan ^{ - 1}}\left( {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)$, then $\dfrac{{dy}}{{dx}}$ is
A. $0$
B. \[\left( {\dfrac{\pi }{4}} \right)\]
C. $1$
D. $\dfrac{1}{2}$

Answer 1

Hint: In the given problem, we are required to differentiate the function $y = {\tan ^{ - 1}}\left( {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)$ with respect to x. Since, $y = {\tan ^{ - 1}}\left( {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)$ is a composite function, so it can be differentiated with help of the chain rule of differentiation. Also, we can simplify the given inverse trigonometric function and then find its derivative with respect to x. We must know the compound angle formula for tangent as \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\].

Complete step by step answer:
So, we have, $y = {\tan ^{ - 1}}\left( {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)$.
We simplify the inverse trigonometric function using the trigonometric formulae.
So, we first divide the numerator and denominator inside the inverse tangent function by cosine. So, we get,
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sin x + \cos x}}{{\cos x}}}}{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}} \right)$
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\tan x + 1}}{{1 - \tan x}}} \right)$
Now, we know the value of $\tan \left( {\dfrac{\pi }{4}} \right)$ is $1$. So, we get,
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\tan x + \tan \left( {\dfrac{\pi }{4}} \right)}}{{1 - \tan x\tan \left( {\dfrac{\pi }{4}} \right)}}} \right)$

Now, we use the compound angle formula for tangent \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\].
$ \Rightarrow y = {\tan ^{ - 1}}\left( {\tan \left( {x + \dfrac{\pi }{4}} \right)} \right)$
The value of ${\tan ^{ - 1}}\left( {\tan y} \right) = y$. So, we get,
$ \Rightarrow y = \left( {x + \dfrac{\pi }{4}} \right)$
Now, we can differentiate the function easily using the power rule of differentiation. So, differentiating both sides of the above equation with respect to x, we get,
$ \therefore \dfrac{{dy}}{{dx}} = 1$
We know that the derivative of a constant is zero.

Hence, the derivative of $y = {\tan ^{ - 1}}\left( {\dfrac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right)$ is $0$.

Note: The given problem may also be solved using the chain rule of differentiation.The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. But we must see ways to simplify the given expression before jumping directly to solve the problem.