Method of DifferentiationmediumFree

Method of Differentiation: Question

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

Question
If y=f ⁣(2x−1x2+1)y=f\!\left(\dfrac{2x-1}{x^{2}+1}\right) and f′(x)=sin⁡x2f'(x)=\sin x^{2}, then y′(0)y'(0) is
Asin⁡1\sin 1
Bcos⁡1\cos 1
C2sin⁡12\sin 1correct
Dcos⁡12\dfrac{\cos 1}{2}
Solution
Step 1: Apply the chain rule:
y′(x)=f′ ⁣(2x−1x2+1)⋅ddx ⁣(2x−1x2+1)y'(x)=f'\!\left(\dfrac{2x-1}{x^{2}+1}\right)\cdot\dfrac{d}{dx}\!\left(\dfrac{2x-1}{x^{2}+1}\right)
 Step 2: Differentiate 2x−1x2+1\dfrac{2x-1}{x^{2}+1} using the quotient rule:
ddx ⁣(2x−1x2+1)=2(x2+1)−(2x−1)(2x)(x2+1)2=2x2+2−4x2+2x(x2+1)2=−2x2+2x+2(x2+1)2\dfrac{d}{dx}\!\left(\dfrac{2x-1}{x^{2}+1}\right)=\dfrac{2(x^{2}+1)-(2x-1)(2x)}{(x^{2}+1)^{2}}=\dfrac{2x^{2}+2-4x^{2}+2x}{(x^{2}+1)^{2}}=\dfrac{-2x^{2}+2x+2}{(x^{2}+1)^{2}}
 Step 3: Since f′(x)=sin⁡x2f'(x)=\sin x^{2}, we have f′ ⁣(2x−1x2+1)=sin⁡ ⁣(2x−1x2+1)2f'\!\left(\dfrac{2x-1}{x^{2}+1}\right)=\sin\!\left(\dfrac{2x-1}{x^{2}+1}\right)^{2}. Step 4: At x=0x=0: 2(0)−10+1=−1\dfrac{2(0)-1}{0+1}=-1 and −0+0+21=2\dfrac{-0+0+2}{1}=2:
y′(0)=sin⁡ ⁣((−1)2)⋅2=sin⁡(1)⋅2=2sin⁡1y'(0)=\sin\!\left((-1)^{2}\right)\cdot 2=\sin(1)\cdot 2=2\sin 1
 Correct answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Method of Differentiation · medium
If y=(sin⁡−1x)2+(cos⁡−1x)2y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2, then (1−x2)d2ydx2−xdydx(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} is:
Method of Differentiation · medium
Let f(x)f(x) and g(x)g(x) be twice differentiable with f′′(x)=g′′(x)f''(x)=g''(x) for all x∈Rx\in\mathbb{R}, f′(1)=2g′(1)=4f'(1)=2g'(1)=4, and g(2)=3f(2)=9g(2)=3f(2)=9. Then f(25)−g(25)f(25)-g(25) equals:
Method of Differentiation · medium
Let ff be a real polynomial of degree nn such that f(x)=f′(x) f′′(x)f(x)=f'(x)\,f''(x) for all x∈Rx\in\mathbb{R}. If f(0)=0f(0)=0, then 36(f′(2)+f′′(2)+∫02f(x) dx)36\left(f'(2)+f''(2)+\displaystyle\int_0^{2}f(x)\,dx\right) is equal to
Method of Differentiation · medium
If f(x)=(ax+b)sin⁡x+(cx+d)cos⁡xf(x) = (ax+b)\sin x+(cx+d)\cos x, then the values of a,b,c,da, b, c, d such that f′(x)=xcos⁡xf'(x) = x\cos x for all xx, then (a+b+c+d)(a+b+c+d) is.

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath