Question

How do you find the derivative of \[\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] ?

Answer 1

Hint:In the above question, we are given a function as \[\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] . We have to find the derivative of the given function. Here arctan is the arc tangent which is also known as the inverse of the tangent. So we can write the function as \[ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] . After than putting \[x = \tan \theta \] in the given function and then by using the trigonometric identity \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] , we can easily find the derivative of the given function.


Complete step by step solution:
Given function is, \[\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right)\]
Since Arctangent can also be written as the inverse of the tangent function.
Hence we can write it as \[ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] .
Now, let us consider
\[ \Rightarrow y = ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)\]
We have to find the derivative of y, i.e \[\dfrac{{dy}}{{dx}}\]
Putting \[x = \tan \theta \] in the equation\[y = ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] , we can write
\[ \Rightarrow y = ta{n^{ - 1}}\left( {\dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)\]
Putting \[1 = \tan \dfrac{\pi }{4}\] , we get
\[ \Rightarrow y = ta{n^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4} \cdot \tan \theta }}} \right)\]
Now using the identity, \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] , we can write
\[ \Rightarrow y = ta{n^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \theta } \right)} \right)\]
Therefore,
\[ \Rightarrow y = \dfrac{\pi }{4} - \theta \]
Since, \[x = \tan \theta \] hence, \[\theta = {\tan ^{ - 1}}x\]
So we can write,
\[ \Rightarrow y = \dfrac{\pi }{4} - {\tan ^{ - 1}}x\]
Differentiating both sides with respect to x, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4}} \right) - \dfrac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}}\]
Since, derivative of a constant is zero and \[\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}\]
Therefore,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{1 + {x^2}}}\]
That is the required derivative of the given function y.
Therefore, the value of derivative of the function \[\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right)\] is \[ - \dfrac{1}{{1 + {x^2}}}\] .

Note:
The given function can also be solved by using the quotient rule and the chain rule.
According to the Quotient rule, if f(x) and g(x) are two functions of x, such that
\[ \Rightarrow y = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}\]
then the derivative of y, \[\dfrac{{dy}}{{dx}}\] or \[y'\] is given by the formula given below
\[ \Rightarrow y' = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{(g(x))}^2}}}\]
And again, if f(x) and g(x) are two functions of x, such that
\[ \Rightarrow y = f\left( x \right) \cdot g\left( x \right)\]
then the derivative of y, \[\dfrac{{dy}}{{dx}}\] or \[y'\] is given by the following formula as
\[ \Rightarrow y' = g(x)f'(x) + f(x)g'(x)\]