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

How do you find the derivative of arctan⁡(1−x1+x)\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right) ?

Explanation

Solution

In the above question, we are given a function as arctan⁡(1−x1+x)\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 tan−1(1−x1+x)ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) . After than putting x=tan⁡θx = \tan \theta in the given function and then by using the trigonometric identity tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\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⁡(1−x1+x)\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 tan−1(1−x1+x)ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) .
Now, let us consider
⇒y=tan−1(1−x1+x)\Rightarrow y = ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)
We have to find the derivative of y, i.e dydx\dfrac{{dy}}{{dx}}
Putting x=tan⁡θx = \tan \theta in the equationy=tan−1(1−x1+x)y = ta{n^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) , we can write
⇒y=tan−1(1−tan⁡θ1+tan⁡θ)\Rightarrow y = ta{n^{ - 1}}\left( {\dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)
Putting 1=tan⁡π41 = \tan \dfrac{\pi }{4} , we get
⇒y=tan−1(tan⁡π4−tan⁡θ1+tan⁡π4⋅tan⁡θ)\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⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} , we can write
⇒y=tan−1(tan⁡(π4−θ))\Rightarrow y = ta{n^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \theta } \right)} \right)
Therefore,
⇒y=π4−θ\Rightarrow y = \dfrac{\pi }{4} - \theta
Since, x=tan⁡θx = \tan \theta hence, θ=tan⁡−1x\theta = {\tan ^{ - 1}}x
So we can write,
⇒y=π4−tan⁡−1x\Rightarrow y = \dfrac{\pi }{4} - {\tan ^{ - 1}}x
Differentiating both sides with respect to x, we get
⇒dydx=ddx(π4)−d(tan⁡−1x)dx\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 ddx(tan⁡−1x)=11+x2\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}
Therefore,
⇒dydx=−11+x2\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⁡(1−x1+x)\arctan \left( {\dfrac{{1 - x}}{{1 + x}}} \right) is −11+x2 - \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
⇒y=f(x)g(x)\Rightarrow y = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}
then the derivative of y, dydx\dfrac{{dy}}{{dx}} or y′y' is given by the formula given below
⇒y′=g(x)f′(x)−f(x)g′(x)(g(x))2\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
⇒y=f(x)⋅g(x)\Rightarrow y = f\left( x \right) \cdot g\left( x \right)
then the derivative of y, dydx\dfrac{{dy}}{{dx}} or y′y' is given by the following formula as
⇒y′=g(x)f′(x)+f(x)g′(x)\Rightarrow y' = g(x)f'(x) + f(x)g'(x)