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Question: If \[y = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}}\] , the...

If y=tan⁡−14x1+5x2+tan⁡−12+3x3−2xy = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}} , then dydx=\dfrac{{dy}}{{dx}} = ?
A. 1(1+25x2)+2(1+x2)\dfrac{1}{{(1 + 25{x^2})}} + \dfrac{2}{{(1 + {x^2})}}
B. 5(1+25x2)+1(1+x2)\dfrac{5}{{(1 + 25{x^2})}} + \dfrac{1}{{(1 + {x^2})}}
C. 5(1+25x2)+1(1+25x2)\dfrac{5}{{(1 + 25{x^2})}} + \dfrac{1}{{(1 + 25{x^2})}}
D. None of these

Explanation

Solution

Here dydx\dfrac{{dy}}{{dx}} represents that we have to find the differentiation of given expression yy with respect to xx . Here we will use formula of ddx(tan⁡−1A)=11+x2\dfrac{d}{{dx}}({\tan ^{ - 1}}A) = \dfrac{1}{{1 + {x^2}}} and tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\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⁡−14x1+5x2+tan⁡−12+3x3−2xy = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}}
Now , taking 4  x4\;x as 5x−x5x - x and for second term we divide numerator and denominator by 33 , on solving we get :
y=tan⁡−15x−x1+5x2+tan⁡−123+x1−23xy = {\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⁡−15x−x1+5x×x+tan⁡−123+x1−23×xy = {\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⁡−1A−tan⁡−1B=tan⁡−1(A−B1+AB){\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right) and tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right), for both the terms , we get ,
y=tan⁡−15x−tan⁡−1x+tan⁡−123+tan⁡−1xy = {\tan ^{ - 1}}5x - {\tan ^{ - 1}}x + {\tan ^{ - 1}}\dfrac{2}{3} + {\tan ^{ - 1}}x ,
Now canceling out the tan⁡−1x{\tan ^{ - 1}}x terms , we get ,
y=tan⁡−15x+tan⁡−123y = {\tan ^{ - 1}}5x + {\tan ^{ - 1}}\dfrac{2}{3}
Now , we use the formula for the differentiation for tan⁡−1A{\tan ^{ - 1}}A , we have ddx(tan⁡−1A)=11+x2\dfrac{d}{{dx}}({\tan ^{ - 1}}A) = \dfrac{1}{{1 + {x^2}}} ,
Therefore , differentiating the expression y=tan⁡−15x+tan⁡−123y = {\tan ^{ - 1}}5x + {\tan ^{ - 1}}\dfrac{2}{3} with respect to xx , we get
dydx=11+25x2×5+0\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 5x5x in the expression , so we have to apply chain rule of derivatives . Therefore , we get
dydx=51+25x2\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⁡−14x1+5x2+tan⁡−12+3x3−2xy = {\tan ^{ - 1}}\dfrac{{4x}}{{1 + 5{x^2}}} + {\tan ^{ - 1}}\dfrac{{2 + 3x}}{{3 - 2x}} as AA and BB use the formula tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){\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 .