The definite integral integral from π/4 to π/4 of f(t) dt has identical upper and lower limits, so its value is identically 0.

JEE MainMathematics

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Differentiating an Integral Equation Using Leibniz's Rule

Q: Why do we multiply by √(x) before differentiating instead of using the quotient rule directly?

Multiplying by √(x) simplifies the Leibniz rule application by avoiding quotient rule fractions, making the substitution at x = π/4 much cleaner.

If φ(x) = 1/(√(x)) integral from π/4 to x of (4√(2)sin t - 3φ'(t))dt, x > 0, then φ'(π/4) is equal to:

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A8/(6+√(π))B4/(6+√(π))C8/(√(π))D4/(6-√(π))

Step-by-step solutionView

Correct answer

Rewriting the relation as $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ and differentiating with respect to $x$ gives $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .

Option analysis

Why each option works or fails

A · 8/(6+√(π))

This is the correct value obtained by using Leibniz's rule and evaluating at $x = π / 4$ . Correctly applying the product rule to $\sqrt{x} ϕ (x)$ gives $\frac{1}{2 \sqrt{x}} ϕ (x) + \sqrt{x} ϕ^{'} (x) = 4 \sqrt{2} \sin x - 3 ϕ^{'} (x)$ . Since $ϕ (π / 4) = 0$ , evaluating at $x = π / 4$ yields $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .

B · 4/(6+√(π))

The student forgets that $4 \sqrt{2} \sin (π / 4) = 4 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4$ , but then multiplies or divides incorrectly by an extra factor of $2$ , leading to a numerator of $4$ instead of $8$ . Compute $(3 + \sqrt{\frac{π}{4}}) ϕ^{'} (\frac{π}{4}) = 4$ , which means $ϕ^{'} (\frac{π}{4}) = \frac{4}{3 + \frac{\sqrt{π}}{2}} = \frac{8}{6 + \sqrt{π}}$ .

C · 8/(√(π))

The student drops the $3 ϕ^{'} (x)$ term when taking the derivative of the integral with respect to $x$ . The integrand contains $- 3 ϕ^{'} (t)$ , which under Leibniz's rule produces $- 3 ϕ^{'} (x)$ ; do not drop terms inside the integrand.

D · 4/(6-√(π))

The student makes a sign error when transposing $- 3 ϕ^{'} (x)$ to the left-hand side, obtaining $3 - \frac{\sqrt{π}}{2}$ instead of $3 + \frac{\sqrt{π}}{2}$ . Adding $3 ϕ^{'} (x)$ to both sides results in $(3 + \sqrt{x}) ϕ^{'} (x)$ , so the denominator must involve $6 + \sqrt{π}$ , not $6 - \sqrt{π}$ .

Reviewed route

Solution

StepWorking

01Given

Given the integral equation: $ϕ (x) = \frac{1}{\sqrt{x}} \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t, x > 0$

02Goal

Find the value of $ϕ^{'} (\frac{π}{4})$ .

03Approach

Clear the denominator by writing $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ . Notice that $ϕ (π / 4) = 0$ because the upper and lower integration limits match. Then, differentiate both sides with respect to $x$ using the product rule and Leibniz's rule, and finally substitute $x = π / 4$ .

04Execute

Rewrite the expression: $\sqrt{x} ϕ (x) = \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ Notice that evaluating at $x = \frac{π}{4}$ gives: $\sqrt{\frac{π}{4}} ϕ (\frac{π}{4}) = \int_{\frac{π}{4}}^{\frac{π}{4}} (\dots) d t = 0 ⟹ ϕ (\frac{π}{4}) = 0$ Differentiating both sides with respect to $x$ using the product rule on the LHS and Leibniz's rule on the RHS: $\frac{1}{2 \sqrt{x}} ϕ (x) + \sqrt{x} ϕ^{'} (x) = 4 \sqrt{2} \sin x - 3 ϕ^{'} (x)$ Group the $ϕ^{'} (x)$ terms: $(\sqrt{x} + 3) ϕ^{'} (x) + \frac{1}{2 \sqrt{x}} ϕ (x) = 4 \sqrt{2} \sin x$

05Execute

Substitute $x = \frac{π}{4}$ into the differentiated equation: $(\sqrt{\frac{π}{4}} + 3) ϕ^{'} (\frac{π}{4}) + \frac{1}{2 \sqrt{π / 4}} ϕ (\frac{π}{4}) = 4 \sqrt{2} \sin (\frac{π}{4})$ Since $ϕ (\frac{π}{4}) = 0$ and $\sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ : $(\frac{\sqrt{π}}{2} + 3) ϕ^{'} (\frac{π}{4}) + 0 = 4 \sqrt{2} (\frac{1}{\sqrt{2}}) = 4$ $(\frac{\sqrt{π} + 6}{2}) ϕ^{'} (\frac{π}{4}) = 4$ $ϕ^{'} (\frac{π}{4}) = \frac{8}{6 + \sqrt{π}}$

✓Verify

Verify the boundary value: evaluating the original equation directly via L'Hopital's rule as $x \to π / 4$ : $\lim_{x \to π / 4} \frac{\int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t}{\sqrt{x}} = \frac{0}{\sqrt{π} / 2} = 0 = ϕ (π / 4)$ . At $x = π / 4$ , $LHS = (\sqrt{π} / 2 + 3) \cdot \frac{8}{6 + \sqrt{π}} = \frac{\sqrt{π} + 6}{2} \cdot \frac{8}{\sqrt{π} + 6} = 4$ , matching $RHS = 4$ .

Hints that build this answer step by step

To eliminate the fraction before differentiating, how can the given equation be rewritten?
$\sqrt{x} ϕ (x) = \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$
What equation results from differentiating both sides of $\sqrt{x} ϕ (x) = \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ with respect to $x$ ?
$\frac{1}{2 \sqrt{x}} ϕ (x) + \sqrt{x} ϕ^{'} (x) = 4 \sqrt{2} \sin x - 3 ϕ^{'} (x)$
Evaluating at $x = \frac{π}{4}$ , what is $ϕ (\frac{π}{4})$ and the resulting equation for $ϕ^{'} (\frac{π}{4})$ ?
$ϕ (\frac{π}{4}) = 0$ , giving $\frac{\sqrt{π}}{2} ϕ^{'} (\frac{π}{4}) + 3 ϕ^{'} (\frac{π}{4}) = 4$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why do we multiply by

\sqrt{x}

before differentiating instead of using the quotient rule directly?

Multiplying by $\sqrt{x}$ simplifies the Leibniz rule application by avoiding quotient rule fractions, making the substitution at $x = π / 4$ much cleaner.

Why is

ϕ (π / 4) = 0

The definite integral $\int_{π / 4}^{π / 4} f (t) d t$ has identical upper and lower limits, so its value is identically 0.

Answer

Rewriting the relation as $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ and differentiating with respect to $x$ gives $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .

Why each option works or fails

A · correct: 8/(6+√(π)) - This is the correct value obtained by using Leibniz's rule and evaluating at $x = π / 4$ . Correctly applying the product rule to $\sqrt{x} ϕ (x)$ gives $\frac{1}{2 \sqrt{x}} ϕ (x) + \sqrt{x} ϕ^{'} (x) = 4 \sqrt{2} \sin x - 3 ϕ^{'} (x)$ . Since $ϕ (π / 4) = 0$ , evaluating at $x = π / 4$ yields $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .
B: 4/(6+√(π)) - The student forgets that $4 \sqrt{2} \sin (π / 4) = 4 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4$ , but then multiplies or divides incorrectly by an extra factor of $2$ , leading to a numerator of $4$ instead of $8$ . Compute $(3 + \sqrt{\frac{π}{4}}) ϕ^{'} (\frac{π}{4}) = 4$ , which means $ϕ^{'} (\frac{π}{4}) = \frac{4}{3 + \frac{\sqrt{π}}{2}} = \frac{8}{6 + \sqrt{π}}$ .
C: 8/(√(π)) - The student drops the $3 ϕ^{'} (x)$ term when taking the derivative of the integral with respect to $x$ . The integrand contains $- 3 ϕ^{'} (t)$ , which under Leibniz's rule produces $- 3 ϕ^{'} (x)$ ; do not drop terms inside the integrand.
D: 4/(6-√(π)) - The student makes a sign error when transposing $- 3 ϕ^{'} (x)$ to the left-hand side, obtaining $3 - \frac{\sqrt{π}}{2}$ instead of $3 + \frac{\sqrt{π}}{2}$ . Adding $3 ϕ^{'} (x)$ to both sides results in $(3 + \sqrt{x}) ϕ^{'} (x)$ , so the denominator must involve $6 + \sqrt{π}$ , not $6 - \sqrt{π}$ .

Step-by-step solution

Given: Given the integral equation: $ϕ (x) = \frac{1}{\sqrt{x}} \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t, x > 0$
Goal: Find the value of $ϕ^{'} (\frac{π}{4})$ .
Approach: Clear the denominator by writing $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ . Notice that $ϕ (π / 4) = 0$ because the upper and lower integration limits match. Then, differentiate both sides with respect to $x$ using the product rule and Leibniz's rule, and finally substitute $x = π / 4$ .
Execute: Rewrite the expression: $\sqrt{x} ϕ (x) = \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ Notice that evaluating at $x = \frac{π}{4}$ gives: $\sqrt{\frac{π}{4}} ϕ (\frac{π}{4}) = \int_{\frac{π}{4}}^{\frac{π}{4}} (\dots) d t = 0 ⟹ ϕ (\frac{π}{4}) = 0$ Differentiating both sides with respect to $x$ using the product rule on the LHS and Leibniz's rule on the RHS: $\frac{1}{2 \sqrt{x}} ϕ (x) + \sqrt{x} ϕ^{'} (x) = 4 \sqrt{2} \sin x - 3 ϕ^{'} (x)$ Group the $ϕ^{'} (x)$ terms: $(\sqrt{x} + 3) ϕ^{'} (x) + \frac{1}{2 \sqrt{x}} ϕ (x) = 4 \sqrt{2} \sin x$
Execute: Substitute $x = \frac{π}{4}$ into the differentiated equation: $(\sqrt{\frac{π}{4}} + 3) ϕ^{'} (\frac{π}{4}) + \frac{1}{2 \sqrt{π / 4}} ϕ (\frac{π}{4}) = 4 \sqrt{2} \sin (\frac{π}{4})$ Since $ϕ (\frac{π}{4}) = 0$ and $\sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ : $(\frac{\sqrt{π}}{2} + 3) ϕ^{'} (\frac{π}{4}) + 0 = 4 \sqrt{2} (\frac{1}{\sqrt{2}}) = 4$ $(\frac{\sqrt{π} + 6}{2}) ϕ^{'} (\frac{π}{4}) = 4$ $ϕ^{'} (\frac{π}{4}) = \frac{8}{6 + \sqrt{π}}$
Verify: Verify the boundary value: evaluating the original equation directly via L'Hopital's rule as $x \to π / 4$ : $\lim_{x \to π / 4} \frac{\int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t}{\sqrt{x}} = \frac{0}{\sqrt{π} / 2} = 0 = ϕ (π / 4)$ . At $x = π / 4$ , $LHS = (\sqrt{π} / 2 + 3) \cdot \frac{8}{6 + \sqrt{π}} = \frac{\sqrt{π} + 6}{2} \cdot \frac{8}{\sqrt{π} + 6} = 4$ , matching $RHS = 4$ .

Differentiating an Integral Equation Using Leibniz's Rule

What feels right?

Rewriting the relation as $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ and differentiating with respect to $x$ gives $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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01Given

02Goal

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Students also ask

Rewriting the relation as $\sqrt{x} ϕ (x) = \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t$ and differentiating with respect to $x$ gives $ϕ^{'} (π / 4) = \frac{8}{6 + \sqrt{π}}$ .