Application of Derivatives — Class 12 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Find the rate of change of the area of a circle with respect to its radius.
A ladder of length 10 m leans against a wall. If the base moves away from the…
Find the maximum and minimum values of f(x) = x³ - 3x on the interval [-2, 2].
Find the equations of the tangent and normal lines to the curve y = x² - 2x a…
The cost function for a company is C(x) = 100 + 5x + 0.1x², where x is the nu…
A particle moves along a line according to s(t) = t³ - 6t² + 9t, where s is i…
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
Find the rate of change of the area of a circle with resp…	✓ Solved
A ladder of length 10 m leans against a wall. If the base…	✓ Solved
Find the maximum and minimum values of f(x) = x³ - 3x on …	✓ Solved
Find the equations of the tangent and normal lines to the…	✓ Solved
The cost function for a company is C(x) = 100 + 5x + 0.1x…	✓ Solved
A particle moves along a line according to s(t) = t³ - 6t…	✓ Solved

Showing 6 of 7 questions

Q1: Find the rate of change of the area of a circle with respect to its radius.

Let A be the area of a circle and r be its radius. We know: A = πr² The rate of change of area with respect to radius is dA/dr. dA/dr = d/dr(πr²) = π × d/dr(r²) = π × 2r = 2πr CONCLUSION: dA/dr = 2πr Interpretation: When the radius is r, a small increase dr in the radius causes an increase of approximately 2πr × dr in the area. This makes geometric sense—2πr is the circumference of the circle.

Q2: A ladder of length 10 m leans against a wall. If the base moves away from the wall at 2 m/s, find the rate at which the top of the ladder slides down the wall when the base is 6 m from the wall.

Let x = distance of base from wall, y = height of top of ladder on the wall. Given: ladder length = 10 m (constant), dx/dt = 2 m/s By Pythagoras: x² + y² = 10² x² + y² = 100 Differentiating both sides with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0 x(dx/dt) + y(dy/dt) = 0 dy/dt = -x(dx/dt)/y When x = 6 m: From x² + y² = 100: 36 + y² = 100 y² = 64 y = 8 m Substituting values: dy/dt = -(6)(2)/8 = -12/8 = -3/2 = -1.5 m/s CONCLUSION: The top of the ladder slides down at 1.5 ...

Q3: Find the maximum and minimum values of f(x) = x³ - 3x on the interval [-2, 2].

Given: f(x) = x³ - 3x on [-2, 2] Step 1: Find critical points f'(x) = 3x² - 3 = 3(x² - 1) = 3(x-1)(x+1) Setting f'(x) = 0: 3(x-1)(x+1) = 0 Critical points: x = 1 and x = -1 Both critical points lie in [-2, 2]. Step 2: Evaluate f at critical points and endpoints f(-2) = (-2)³ - 3(-2) = -8 + 6 = -2 f(-1) = (-1)³ - 3(-1) = -1 + 3 = 2 f(1) = (1)³ - 3(1) = 1 - 3 = -2 f(2) = (2)³ - 3(2) = 8 - 6 = 2 Step 3: Identify maximum and minimum Maximum value: 2 (occurs at x = -1 and x = 2) Minimum value: -...

Q4: Find the equations of the tangent and normal lines to the curve y = x² - 2x at the point (1, -1).

Given: y = x² - 2x, point (1, -1) Step 1: Verify the point lies on the curve y(1) = 1² - 2(1) = 1 - 2 = -1 ✓ Step 2: Find the slope of the tangent dy/dx = 2x - 2 At x = 1: dy/dx = 2(1) - 2 = 0 Slope of tangent = 0 Step 3: Find equation of tangent line Using point-slope form: y - y₁ = m(x - x₁) y - (-1) = 0(x - 1) y + 1 = 0 y = -1 Tangent line: y = -1 (horizontal line) Step 4: Find equation of normal line Slope of normal = -1/(slope of tangent) = -1/0 = undefined A normal with undefined slo...

Q5: The cost function for a company is C(x) = 100 + 5x + 0.1x², where x is the number of units. Find the marginal cost when x = 10.

Given: C(x) = 100 + 5x + 0.1x² The marginal cost is the rate of change of cost with respect to units produced, which is dC/dx. Step 1: Find the marginal cost function dC/dx = d/dx(100 + 5x + 0.1x²) = 0 + 5 + 0.1(2x) = 5 + 0.2x Step 2: Evaluate marginal cost at x = 10 Marginal cost at x = 10 = 5 + 0.2(10) = 5 + 2 = 7 CONCLUSION: The marginal cost when x = 10 is 7 (currency units per unit produced). Interpretation: When the company is ...

Q6: A particle moves along a line according to s(t) = t³ - 6t² + 9t, where s is in metres and t is in seconds. Find the velocity and acceleration when t = 2.

Given: s(t) = t³ - 6t² + 9t Step 1: Find velocity Velocity v(t) = ds/dt v(t) = d/dt(t³ - 6t² + 9t) = 3t² - 12t + 9 At t = 2: v(2) = 3(2)² - 12(2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3 m/s Step 2: Find acceleration Acceleration a(t) = dv/dt = d²s/dt² a(t) = d/dt(3t² - 12t + 9) = 6t - 12 At t = 2: a(2) = 6(2) - 12 = 12 - 12 = 0 m/s² CONCLUSION: At t = 2: - Velocity = -3 m/s (moving in negative direction) - Acceleration = 0 m/s² (no change in velocity at thi...

Showing 6 of 7 questions. Visit the full page for complete solutions.