Question

If \[y = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}}\] , then \[\dfrac{{dy}}{{dx}} = \]?
A. \[\dfrac{1}{{(1 + 25{x^2})}} + \dfrac{2}{{(1 + {x^2})}}\]
B. \[\dfrac{5}{{(1 + 25{x^2})}} + \dfrac{1}{{(1 + {x^2})}}\]
C. \[\dfrac{5}{{(1 + 25{x^2})}} + \dfrac{1}{{(1 + 25{x^2})}}\]
D. None of these

Answer 1

Hint: Here \[\dfrac{{dy}}{{dx}}\] represents that we have to find the differentiation of given expression \[y\] with respect to \[x\] . Here we will use formula of \[\dfrac{d}{{dx}}({\tan ^{ - 1}}A) = \dfrac{1}{{1 + {x^2}}}\] and \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\] for solving this .

Complete step by step answer:
Given : - \[y = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}}\]
Now , taking \[4\;x\] as \[5x - x\] and for second term we divide numerator and denominator by \[3\] , on solving we get :
\[y = {\tan ^{ - 1}}\dfrac{{5x - x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{\dfrac{2}{3} + x}}{{1 - \dfrac{2}{3}x}}\] ….. eqn (a)
Further rewriting the equation ,
\[y = {\tan ^{ - 1}}\dfrac{{5x - x}}{{1 + 5x \times x}} + {\tan ^{ - 1}}\dfrac{{\dfrac{2}{3} + x}}{{1 - \dfrac{2}{3} \times x}}\] ,
Now on comparing eqn (a) with the formula \[{\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right)\] and \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\], for both the terms , we get ,
\[y = {\tan ^{ - 1}}5x - {\tan ^{ - 1}}x + {\tan ^{ - 1}}\dfrac{2}{3} + {\tan ^{ - 1}}x\] ,
Now canceling out the \[{\tan ^{ - 1}}x\] terms , we get ,
\[y = {\tan ^{ - 1}}5x + {\tan ^{ - 1}}\dfrac{2}{3}\]
Now , we use the formula for the differentiation for \[{\tan ^{ - 1}}A\] , we have \[\dfrac{d}{{dx}}({\tan ^{ - 1}}A) = \dfrac{1}{{1 + {x^2}}}\] ,
Therefore , differentiating the expression \[y = {\tan ^{ - 1}}5x + {\tan ^{ - 1}}\dfrac{2}{3}\] with respect to \[x\] , we get
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + 25{x^2}}} \times 5 + 0\] , we get zero for second term as it is a constant terms , so differentiation of any constant will always be zero . Also , we have multiplied by as it is \[5x\] in the expression , so we have to apply chain rule of derivatives . Therefore , we get
\[\dfrac{{dy}}{{dx}} = \dfrac{5}{{1 + 25{x^2}}}\]

So, the correct answer is “Option A”.

Note: Alternate method :
We can assume both the terms of expression as \[y = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}}\] as \[A\] and \[B\] use the formula \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]. But this method will be complicated and try to avoid it .
We have to remember the for trigonometric identities such as the \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\] and differentiation of trigonometric ratios along with the chain rule as sometimes different expression are given and your solution will be consider wrong .