Definite IntegrationeasyFree

Derivative of f(x)=∫ from 1/x to √x of cos t² dt | JEE

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

Question

f(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt

for

x>0

, then

\dfrac{df(x)}{dx}

\dfrac{\sqrt{x}\cos x+2\cos\!\big(x^{-2}\big)}{2x\sqrt{x}}

\dfrac{x\sqrt{x}\cos x+2\cos\!\big(x^{-2}\big)}{2x^{2}}

correct

2\sqrt2\cos x-\dfrac{2}{x}\cos\dfrac{1}{x}

DNone of these

Solution

Step 1: Apply Leibniz's rule for differentiating an integral with variable limits:

f'(x)=\cos\!\big((\sqrt{x})^{2}\big)\dfrac{d}{dx}\sqrt{x}-\cos\!\Big(\big(\tfrac1x\big)^{2}\Big)\dfrac{d}{dx}\Big(\dfrac1x\Big).

Step 2: Compute the two endpoint contributions. For the upper limit,

\cos\!\big((\sqrt{x})^{2}\big)=\cos x

and

\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}

. For the lower limit,

\cos\!\Big(\dfrac{1}{x^{2}}\Big)

and

\dfrac{d}{dx}\Big(\dfrac1x\Big)=-\dfrac{1}{x^{2}}

. Step 3: Combine, watching the sign on the lower limit:

f'(x)=\dfrac{\cos x}{2\sqrt{x}}-\cos\!\Big(\dfrac{1}{x^{2}}\Big)\!\left(-\dfrac{1}{x^{2}}\right)=\dfrac{\cos x}{2\sqrt{x}}+\dfrac{\cos\!\big(x^{-2}\big)}{x^{2}}.

Step 4: Put over the common denominator

2x^{2}

. Since

\dfrac{\cos x}{2\sqrt{x}}=\dfrac{x\sqrt{x}\cos x}{2x^{2}}

and

\dfrac{\cos(x^{-2})}{x^{2}}=\dfrac{2\cos(x^{-2})}{2x^{2}}

f'(x)=\dfrac{x\sqrt{x}\cos x+2\cos\!\big(x^{-2}\big)}{2x^{2}}.

Correct answer: (2)

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