JEE Main Differentiation Important Questions - Are you preparing to take the JEE Main 2024 Exam? 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 2024 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. 

The National Testing Agency or NTA has released the JEE Main 2024 Exam Dates on the official website. JEE Main 2024 Exam date session 1 is January 24 to February 1, 2024, and the JEE Main session 2 exam date is April 1 to 15, 2024. Candidates can check the JEE Main Differentiation Important Questions below. 
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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 whilst taking the JEE Main 2024 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 pace 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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