JEE Main Differentiation Important Questions

JEE Main Differentiation Important Questions - Are you preparing to take the JEE Main 2026 ? Do you want to know how to pass the JEE Main Differentiation Section? We've got your back. By preparing and answering the Differentiation section of the JEE Main Mathematics Differentiation Questions, you may ace the JEE Main 2026 exam. To tackle the differentiation section of the question paper, one must have a thorough comprehension of the concepts of differentiation as well as effective problem-solving skills.

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JEE Main Differentiation Important Questions

Solving questions related to differentiation in the JEE Mains mathematics portion is a difficult challenge. Some essential questions, along with their step-by-step solutions, are provided in this article for students to prepare and ace their JEE Main. These questions will also aid you in time management while taking the JEE Main 2026 Exam.

Question 1: If f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)] and its first derivative with respect to x is (- b / a) log 2 when x = 1, where a and b are integers, then the minimum value of |a2 – b2| is:

Solution:

Given that f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)]

cos-1 [1 – 4x] / [1 + 4x]

Let 2x = t > 0

cos-1 [(1 – t2) / (1 + t2)], t > 0 and t = tan θ

cos-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)

⇒ 2θ

sin {cos-1 [[1 – 4x] / [1 + 4x]]} = sin2θ

So y = [2 tanθ / (1 + tan2 θ)]

= 2t / (1 + t2)

= [2 × 2x] / [1 + 4x]

dy/dx = [20 ln 2 – 32 ln 2] / 25

= – 12 ln 2 / 25

a = 25, b = 12

|a2 – b2|min = |252 – 122| = 481

Hence, the minimum value of |a2 – b2| is 481.

Question 2 : Find y(π) if y(0) = 7 and dy/dx = 2(y – 2 sin x – 10)x + 2 cos x.

Solution:

Given that dy/dx = 2(y – 2 sin x – 10)x + 2 cos x

dy/dx – 2 cos x = 2(y – 2 sin x – 10)x

(d/dx (y – 2 sin x – 10))/((y – 2 sin x – 10) ) = 2x

⇒∫d(y – 2 sin x – 10)/((y – 2 sin x – 10) ) = ∫2x dx

⇒ log |y – 2 sin x – 10| = x2 + C

When x = 0, y = 7

⇒ log |7 – 0 – 10| = 0 + C

So C = log 3

When x = π

⇒ log |y – 2 sin π – 10| = π2+ log 3

⇒ log ((y – 10)/3) = π

= y( π ) = 3e

Question 3: d/dx ( log e x) ( log a x) =

(a) (1/x) log a x

(b) (1/x) log x x

(d) (2/x) log a x

(d) (2/x) log x

Solution:

Let y = ( loge x) ( log a x)

= (log x/log e) ( log x/ log a)

= (log x)2/log a

Differentiating with respect to x, we get;

dy/dx = 2 log x (1/x)/log a

= (2/x) log a x

Hence, option (c) is the answer.

Question 4: If f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)] and its first derivative with respect to x is (- b / a) loge 2 when x = 1, where a and b are integers, then the minimum value of |a2 – b2| is:

Solution:

Given that f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)]

cos-1 [1 – 4x] / [1 + 4x]

Let 2x = t > 0

cos-1 [(1 – t2) / (1 + t2)], t > 0 and t = tan θ

cos-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)

⇒ 2θ

sin {cos-1 [[1 – 4x] / [1 + 4x]]} = sin2θ

So y = [2 tanθ / (1 + tan2 θ)]

= 2t / (1 + t2)

= [2 × 2x] / [1 + 4x]

dy/dx = [20 ln 2 – 32 ln 2] / 25

= – 12 ln 2 / 25

a = 25, b = 12

|a2 – b2|min = |252 – 122| = 481

Hence, the minimum value of |a2 – b2| is 481.

Question 5 : If ey + xy = e, then the value of d2y/dx2 for x = 0 is

(a) 1/e

(b) 1/e2

(c) 1/e3

(d) none of these

Solution:

Given that e y + xy = e

When x = 0, we get y = 1

Differentiate w.r.t.x

eydy/dx + x dy/dx + y = 0 …(i)

Put x = 0 and y = 1, we get

dy/dx = -y/(x+ ey) = -1/e

Again differentiate (i) w.r.t.x

ey(dy/dx)2 + eyd2y/dx2 + dy/dx + x d2y/dx2 + dy/dx = 0

d2y/dx2 [ ey + x] = -2dy/dx – ey(dy/dx)2 …(ii)

Put x = 0 and y = 1 and dy/dx = -1/e in (ii), we get;

d2y/dx2 [ e + 0] = 2/e – e/e2

d2y/dx2 e = 1/e

d2y/dx2 = 1/e2

Hence, option (b) is the answer.

Question 6 : If loge (x+y) = 4xy, find (d2y)/(dx2) at x = 0.

Solution:

Given that loge (x+y) = 4xy

Differentiating with respect to x, we get-

(1/(x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]

1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)

If x = 0, then y = 1.

From (i), we get

1 + dy/dx = 4

⇒ dy/dx = 3

Again differentiate (i) w.r.t. x.

d2y/dx2 = 4(x + y)[x (d2y)/(dx2) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))

At x = 0, y = 1, dy/dx = 3

d2y/dx2 = 4(0 + 1)[0 + 2x3]+4[0 + 1](1 + 3)

= 40

So, d2y/dx2 = 40.

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About Differentiation

In calculus, the process of determining a function's derivatives is known as differentiation. A derivative is the rate at which a function changes in relation to another quantity. Sir Isaac Newton established the laws of Differential Calculus. Limit and derivative ideas are applied in many scientific areas. Calculus' main principles are differentiation and integration.

Differentiation determines the highest or lowest value of a function, the velocity and acceleration of moving objects, and the tangent of a curve. If y = f(x) and x is differentiable, the differentiation is denoted by f'(x) or dy/dx.

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