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Question: Find the differentiation of y w.r.t to x, if \(y = {\cot ^{ - 1}}\dfrac{{1 - x}}{{1 + x}}\)?...

Find the differentiation of y w.r.t to x, if y=cot⁡−11−x1+xy = {\cot ^{ - 1}}\dfrac{{1 - x}}{{1 + x}}?

Explanation

Solution

We can substitute and rearrange them to make it solved easily. We can substitute tan⁡a=x\tan a=x, and tan⁡π4=1\tan \dfrac{\pi }{4} = 1. We also know that cot⁡−1cot⁡a=a{\cot ^{ - 1}}\cot a = a. then by substituting the value of a, and doing the differentiation.
We should try to substitute our term such that all trigonometric functions get exempted from function.

Complete step-by-step solution:
we have been given that y=cot⁡−11−x1+xy = {\cot ^{ - 1}}\dfrac{{1 - x}}{{1 + x}}
we can see that we can make the above equation in simpler form. If we want to differentiate in simpler form, we should cancel the cot⁡−1{\cot ^{ - 1}}. So, for removing cot⁡−1{\cot ^{ - 1}}, we have to substitute x=tan⁡ax = \tan aand1=tan⁡45∘1 = \tan {45^ \circ }
y=cot⁡−11−tan⁡a1+tan⁡ay = {\cot ^{ - 1}}\dfrac{{1 - \tan a}}{{1 + \tan a}}
We will use formula tan⁡(a−b)=(tan⁡a−tan⁡b1+tan⁡a×tan⁡b)\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}} \right) to simplify,
⇒y=cot⁡−1tan⁡(π4−a)\Rightarrow y = {\cot ^{ - 1}}\tan \left( {\dfrac{\pi }{4} - a} \right)
\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left\\{ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} - a} \right)} \right\\}} \right]
\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left\\{ {\dfrac{\pi }{4} + a} \right\\}} \right]
So here cot⁡−1{\cot ^{ - 1}}and cot⁡\cot . So now substituting we get,
\Rightarrow y = \left\\{ {\dfrac{\pi }{4} + a} \right\\}
We have assumed tana=x .so, a=tan⁡−1xa = {\tan ^{ - 1}}x
\Rightarrow y = \left\\{ {\dfrac{\pi }{4} + {{\tan }^{ - 1}}x} \right\\}
Now we know$$$$$${\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}},thenthedifferentiationbecomes, then the differentiation becomes \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\{ {\dfrac{\pi }{4} + {{\tan }^{ - 1}}x} \right\}Wewilldifferentiateseparately,weget We will differentiate separately, we get \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4}} \right) + \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) \Rightarrow \dfrac{{dy}}{{dx}} = 0 + \dfrac{1}{{1 + {x^2}}} **So, the derivative of $y = {\cot ^{ - 1}}\dfrac{{1 - x}}{{1 + x}}$ is\dfrac{1}{{1 + {x^2}}}$$ .**

Note: We should be familiar with the properties and identities such as tan⁡(a−b)=(tan⁡a−tan⁡b1+tan⁡a×tan⁡b)\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}} \right)$$$$ . We will take the value of 1 as   tanπ4\;tan\dfrac{\pi }{4} is also a smart move, because it makes our calculation easy. Take care of differentiation of tan⁡−1θ{\tan ^{ - 1}}\theta .We should take utmost care for where to substitute what, and it would come with a lot of practice.