Class 12 — Mathematics (NCERT)

Application of Derivatives

Class 12

  • ✓By the end of this lesson students will be able to understand the concept of rate of change of quantities.
  • ✓By the end of this lesson students will be able to determine the equations of tangents and normals to a curve at a given point.
  • ✓By the end of this lesson students will be able to find the intervals in which a function is increasing or decreasing.
  • ✓By the end of this lesson students will be able to find the local maxima and local minima of a function using both first and second derivative tests.
  • ✓By the end of this lesson students will be able to solve practical problems involving maxima and minima.

Key concepts

Rate of Change of Quantities

If a quantity y varies with another quantity x, i.e., y = f(x), then dy/dx represents the rate of change of y with respect to x. Similarly, if a quantity y varies with time t, then dy/dt represents the rate of change of y with respect to time t. These problems often involve relating the rates of change of two or more related quantities using the chain rule.

dy/dx, dA/dt, dV/dt (where A is area, V is volume, t is time)
Tangents and Normals

The derivative dy/dx at a point (x₀, y₀) on a curve y = f(x) gives the slope of the tangent to the curve at that point. The normal to the curve at that point is a line perpendicular to the tangent. \n\nIf the slope of the tangent is m_T, then the slope of the normal is m_N = -1/m_T (provided m_T ≠ 0). \n\nSpecial cases: \n1. If m_T = 0, the tangent is horizontal (equation y = y₀), and the normal is vertical (equation x = x₀). \n2. If m_T is undefined (i.e., dx/dy = 0), the tangent is vertical (equation x = x₀), and the normal is horizontal (equation y = y₀).

Slope of tangent, m_T = (dy/dx) at (x₀, y₀)\nEquation of tangent: y - y₀ = m_T(x - x₀)\nSlope of normal, m_N = -1/m_T\nEquation of normal: y - y₀ = m_N(x - x₀)
Increasing and Decreasing Functions

A function f(x) is said to be increasing on an interval (a, b) if for any x₁, x₂ ∈ (a, b), x₁ < x₂ implies f(x₁) ≤ f(x₂). It is strictly increasing if f(x₁) < f(x₂). \n\nA function f(x) is said to be decreasing on an interval (a, b) if for any x₁, x₂ ∈ (a, b), x₁ < x₂ implies f(x₁) ≥ f(x₂). It is strictly decreasing if f(x₁) > f(x₂). \n\nTest for increasing/decreasing functions: \n1. If f'(x) > 0 for all x in (a, b), then f(x) is strictly increasing on (a, b). \n2. If f'(x) < 0 for all x in (a, b), then f(x) is strictly decreasing on (a, b). \n3. If f'(x) = 0 for all x in (a, b), then f(x) is a constant function on (a, b).

f'(x) > 0 (strictly increasing)\nf'(x) < 0 (strictly decreasing)
Maxima and Minima (Local/Relative)

A function f(x) has a local maximum at a point c if f(c) is the greatest value of f in some open interval containing c. Similarly, f(x) has a local minimum at c if f(c) is the least value of f in some open interval containing c. Points where f'(x) = 0 or f'(x) is undefined are called critical points.\n\nFirst Derivative Test: \n1. If f'(x) changes sign from positive to negative as x increases through c, then c is a point of local maximum. \n2. If f'(x) changes sign from negative to positive as x increases through c, then c is a point of local minimum. \n3. If f'(x) does not change sign as x increases through c, then c is neither a local maximum nor a local minimum (it's a point of inflection).\n\nSecond Derivative Test: \n1. Find f'(x) and f''(x). \n2. Find critical points by setting f'(x) = 0. Let c be a critical point. \n3. If f''(c) < 0, then c is a point of local maximum. \n4. If f''(c) > 0, then c is a point of local minimum. \n5. If f''(c) = 0, the test fails. Use the first derivative test.

First Derivative Test: Sign change of f'(x)\nSecond Derivative Test: f''(c) < 0 (local max), f''(c) > 0 (local min)
Absolute Maxima and Minima

For a continuous function f(x) on a closed interval [a, b], the absolute maximum (or global maximum) is the largest value of f(x) on that interval, and the absolute minimum (or global minimum) is the smallest value. \n\nSteps to find absolute maxima/minima: \n1. Find all critical points of f(x) in the open interval (a, b). \n2. Evaluate f(x) at these critical points and at the endpoints a and b. \n3. The largest value among these is the absolute maximum, and the smallest value is the absolute minimum.

Compare f(c) for critical points c ∈ (a,b) and f(a), f(b).

Key facts to remember

  • 1dy/dx represents the instantaneous rate of change of y with respect to x.
  • 2The slope of the tangent to a curve y = f(x) at (x₀, y₀) is given by (dy/dx) at (x₀, y₀).
  • 3The slope of the normal to a curve at a point is the negative reciprocal of the slope of the tangent at that point (if the tangent is not horizontal or vertical).
  • 4A function f(x) is strictly increasing if f'(x) > 0 and strictly decreasing if f'(x) < 0.
  • 5Critical points are points where f'(x) = 0 or f'(x) is undefined.
  • 6For a local maximum at c, f'(x) changes from positive to negative, or f''(c) < 0.
  • 7For a local minimum at c, f'(x) changes from negative to positive, or f''(c) > 0.
  • 8To find absolute maxima/minima on a closed interval [a, b], evaluate f(x) at critical points in (a, b) and at the endpoints a and b.

Worked examples

Example 1

The volume of a cube is increasing at the rate of 9 cm³/s. How fast is the surface area increasing when the length of an edge is 10 cm?

ILet x be the length of an edge of the cube, V be its volume, and S be its surface area.
IIGiven: dV/dt = 9 cm³/s. We need to find dS/dt when x = 10 cm.
IIIThe volume of a cube is V = x³.
IVDifferentiating V with respect to time t, we get:
VdV/dt = d/dt (x³) = 3x² (dx/dt) (using chain rule)
VISubstitute the given dV/dt = 9:
VII9 = 3x² (dx/dt)
VIIITherefore, dx/dt = 9 / (3x²) = 3/x² ...(1)
9The surface area of a cube is S = 6x².
10Differentiating S with respect to time t, we get:
11dS/dt = d/dt (6x²) = 12x (dx/dt) (using chain rule)
12Substitute the value of dx/dt from (1) into this equation:
13dS/dt = 12x (3/x²) = 36/x
14Now, we need to find dS/dt when x = 10 cm:
15dS/dt |_(x=10) = 36/10 = 3.6 cm²/s

Answer

The surface area is increasing at the rate of 3.6 cm²/s.

Remember to use the chain rule when differentiating quantities with respect to time.

Example 2

Find the equations of the tangent and normal to the curve y = x³ - 3x² - 9x + 7 at the point where x = 3.

IFirst, find the y-coordinate of the point when x = 3:
IIy = (3)³ - 3(3)² - 9(3) + 7
IIIy = 27 - 3(9) - 27 + 7
IVy = 27 - 27 - 27 + 7 = -20
VSo, the point is (3, -20).
VINext, find the derivative dy/dx to get the slope of the tangent:
VIIdy/dx = d/dx (x³ - 3x² - 9x + 7)
VIIIdy/dx = 3x² - 6x - 9
9Now, calculate the slope of the tangent (m_T) at x = 3:
10m_T = (dy/dx) |_(x=3) = 3(3)² - 6(3) - 9
11m_T = 3(9) - 18 - 9
12m_T = 27 - 18 - 9 = 0
13Since the slope of the tangent is 0, the tangent is a horizontal line.
14Equation of the tangent: y - y₀ = m_T(x - x₀)
15y - (-20) = 0(x - 3)
16y + 20 = 0
17y = -20
18For the normal, since the tangent is horizontal, the normal must be a vertical line.
19The equation of a vertical line passing through (x₀, y₀) is x = x₀.
20Equation of the normal: x = 3

Answer

Equation of the tangent is y = -20. Equation of the normal is x = 3.

A horizontal tangent has slope 0, and its normal is a vertical line. A vertical tangent has undefined slope, and its normal is a horizontal line.

Example 3

Find the local maximum and local minimum values of the function f(x) = x³ - 6x² + 9x + 15.

IFirst, find the first derivative of the function:
IIf'(x) = d/dx (x³ - 6x² + 9x + 15)
IIIf'(x) = 3x² - 12x + 9
IVTo find critical points, set f'(x) = 0:
V3x² - 12x + 9 = 0
VIDivide by 3: x² - 4x + 3 = 0
VIIFactor the quadratic: (x - 1)(x - 3) = 0
VIIISo, the critical points are x = 1 and x = 3.
9Now, find the second derivative of the function:
10f''(x) = d/dx (3x² - 12x + 9)
11f''(x) = 6x - 12
12Apply the Second Derivative Test at each critical point:
13At x = 1:
14f''(1) = 6(1) - 12 = 6 - 12 = -6
15Since f''(1) < 0, x = 1 is a point of local maximum.
16Local maximum value: f(1) = (1)³ - 6(1)² + 9(1) + 15 = 1 - 6 + 9 + 15 = 19.
17At x = 3:
18f''(3) = 6(3) - 12 = 18 - 12 = 6
19Since f''(3) > 0, x = 3 is a point of local minimum.
20Local minimum value: f(3) = (3)³ - 6(3)² + 9(3) + 15 = 27 - 54 + 27 + 15 = 15.

Answer

The local maximum value is 19 at x = 1. The local minimum value is 15 at x = 3.

If f''(c) = 0, the second derivative test fails, and you must use the first derivative test.

Common mistakes

  • ✗Confusing the slope of the tangent with the slope of the normal, or forgetting the negative reciprocal relationship.
  • ✗Incorrectly applying the chain rule in related rates problems, especially when differentiating with respect to time.
  • ✗Forgetting to check the endpoints of the interval when finding absolute maxima and minima.
  • ✗Misinterpreting the sign change in the first derivative test (e.g., positive to negative for minimum).
  • ✗Algebraic errors in differentiation, solving for critical points, or evaluating function values.

Exam tips

  • ★Read the question carefully to identify what quantity needs to be maximised/minimised or what rate of change is required.
  • ★For application problems, draw a clear diagram and label all variables. Formulate the problem mathematically.
  • ★Clearly state which test (First Derivative Test or Second Derivative Test) you are using for maxima/minima and show all steps.
  • ★Always write down the units for rates of change or final answers in application problems.

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