Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If y=tan⁡−14x1+5x2+tan⁡−12+3x3−2xy = \tan^{-1}\dfrac{4x}{1+5x^2}+\tan^{-1}\dfrac{2+3x}{3-2x} and dydx=k1+25x2\dfrac{dy}{dx} = \dfrac{k}{1+25x^2}, then kk is.
Solution
Answer: 5
Step 1: Decompose the first term
tan⁡−14x1+5x2=tan⁡−15x−x1+5x⋅x=tan⁡−15x−tan⁡−1x\tan^{-1}\frac{4x}{1+5x^2} = \tan^{-1}\frac{5x-x}{1+5x\cdot x} = \tan^{-1}5x - \tan^{-1}x
 Step 2: Decompose the second term
tan⁡−12+3x3−2x=tan⁡−123+x1−23x=tan⁡−123+tan⁡−1x\tan^{-1}\frac{2+3x}{3-2x} = \tan^{-1}\frac{\frac{2}{3}+x}{1-\frac{2}{3}x} = \tan^{-1}\frac{2}{3}+\tan^{-1}x
 Step 3: Combine and differentiate
y=tan⁡−15x−tan⁡−1x+tan⁡−123+tan⁡−1x=tan⁡−15x+tan⁡−123y = \tan^{-1}5x - \tan^{-1}x + \tan^{-1}\frac{2}{3}+\tan^{-1}x = \tan^{-1}5x+\tan^{-1}\frac{2}{3}
dydx=51+25x2  ⟹  k=5\frac{dy}{dx} = \frac{5}{1+25x^2} \implies k = 5
 Answer: 5
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