Method of DifferentiationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Derivative of Inverse Trig Sum at x=√3/2: 1 | JEE Main 2026

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

Question
If y=tan1(3cosx4sinx4cosx+3sinx)+2tan1(x1+1x2)y=\tan^{-1}\left(\dfrac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)+2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1-x^2}}\right), then dydx\dfrac{dy}{dx} at x=32x=\dfrac{\sqrt3}{2} is equal to
A33
B1-1
C11correct
D22
Solution
Step 1: Divide numerator and denominator of the first fraction by 4cosx4\cos x:
3cosx4sinx4cosx+3sinx=34tanx1+34tanx=tan(tan134x).\frac{3\cos x-4\sin x}{4\cos x+3\sin x}=\frac{\frac34-\tan x}{1+\frac34\tan x}=\tan\left(\tan^{-1}\tfrac34-x\right).
So the first term is tan134x\tan^{-1}\dfrac34-x, whose derivative is 1-1. Step 2: For the second term, put x=sinθx=\sin\theta:
2tan1(sinθ1+cosθ)=2tan1(tanθ2)=θ=sin1x,2\tan^{-1}\left(\frac{\sin\theta}{1+\cos\theta}\right)=2\tan^{-1}\left(\tan\tfrac{\theta}{2}\right)=\theta=\sin^{-1}x,
whose derivative is 11x2\dfrac{1}{\sqrt{1-x^2}}. Step 3: So dydx=1+11x2\dfrac{dy}{dx}=-1+\dfrac{1}{\sqrt{1-x^2}}. At x=32x=\dfrac{\sqrt3}{2}, 1x2=12\sqrt{1-x^2}=\dfrac12:
dydx=1+2=1.\frac{dy}{dx}=-1+2=1.
Correct answer: (3)
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