Question

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

Answer 1

Hint: We can substitute and rearrange them to make it solved easily. We can substitute $\tan a=x$, and \[\tan \dfrac{\pi }{4} = 1\]. We also know that \[{\cot ^{ - 1}}\cot a = a\]. 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 ^{ - 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 ${\cot ^{ - 1}}$. So, for removing ${\cot ^{ - 1}}$, we have to substitute $x = \tan a$and$1 = \tan {45^ \circ }$
$y = {\cot ^{ - 1}}\dfrac{{1 - \tan a}}{{1 + \tan a}}$
We will use formula $\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}} \right)$ to simplify,
$\Rightarrow y = {\cot ^{ - 1}}\tan \left( {\dfrac{\pi }{4} - a} \right)$
$\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left\{ {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} - a} \right)} \right\}} \right]$
$\Rightarrow y = {\cot ^{ - 1}}\left[ {\cot \left\{ {\dfrac{\pi }{4} + a} \right\}} \right]$
So here ${\cot ^{ - 1}}$and $\cot $. So now substituting we get,
$\Rightarrow y = \left\{ {\dfrac{\pi }{4} + a} \right\}$
We have assumed tana=x .so, \[a = {\tan ^{ - 1}}x\]
$\Rightarrow y = \left\{ {\dfrac{\pi }{4} + {{\tan }^{ - 1}}x} \right\}$
Now we know\[\]\[{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}\] , then the differentiation becomes
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\{ {\dfrac{\pi }{4} + {{\tan }^{ - 1}}x} \right\}\]
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 $\tan \left( {a - b} \right) = \left( {\dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}} \right)$\[\] . We will take the value of 1 as \[\;tan\dfrac{\pi }{4}\] is also a smart move, because it makes our calculation easy. Take care of differentiation of \[{\tan ^{ - 1}}\theta \].We should take utmost care for where to substitute what, and it would come with a lot of practice.