Application of Derivatives

In this chapter: rate of change of quantities, increasing and decreasing functions, critical points, local maxima and minima by the first and second derivative tests, absolute maximum and minimum on a closed interval, optimisation problems, and equations of tangents and normals.

Rate of change

dy/dx is the rate of change of y with respect to x, and (dy/dx)_{x = x₀} is that rate at the particular value x₀. When both x and y depend on time t, the chain rule links their rates: dy/dt = (dy/dx)(dx/dt). Most "related rates" questions are solved in three steps: write the formula connecting the quantities, differentiate with respect to t, and only then substitute the given values.

Worked example: The radius of a circular ripple increases at 3 cm/s. How fast is the enclosed area increasing when the radius is 10 cm?
Solution: A = πr², so dA/dt = 2πr (dr/dt) = 2π(10)(3) = 60π cm²/s. Substituting r = 10 before differentiating would give A = 100π, a constant, and a rate of zero, which is why the order of steps matters.

Increasing and decreasing functions

f is increasing on an interval I if x₁ < x₂ implies f(x₁) ≤ f(x₂), and strictly increasing if it implies f(x₁) < f(x₂). Decreasing is defined the same way with the inequalities reversed. The derivative test (for f continuous on [a, b] and differentiable on (a, b)):

• f′(x) > 0 for every x in (a, b) ⇒ f is strictly increasing on [a, b].
• f′(x) < 0 for every x in (a, b) ⇒ f is strictly decreasing on [a, b].
• f′(x) = 0 for every x in (a, b) ⇒ f is constant on [a, b].

The converse of the first statement is not quite true: f(x) = x³ is strictly increasing on R, yet f′(0) = 0. An isolated zero of f′ does not spoil monotonicity.

Method: find f′(x), factorise it, mark its zeros on a number line and check the sign of f′ in each interval.

Maxima and minima

f has a local maximum at c if f(c) ≥ f(x) for all x in some open interval around c, and a local minimum if f(c) ≤ f(x) there. A point c where f′(c) = 0 or f is not differentiable is a critical point. Local extrema of a function can occur only at critical points, but not every critical point is an extremum (x³ at 0 is a point of inflection).

Worked example: Find the intervals of increase and decrease and the local extrema of f(x) = 2x³ − 9x² + 12x + 15.
Solution: f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2). f′ > 0 for x < 1, f′ < 0 for 1 < x < 2, and f′ > 0 for x > 2. So f is increasing on (−∞, 1], decreasing on [1, 2] and increasing on [2, ∞). f″(x) = 12x − 18: f″(1) = −6 < 0, so x = 1 gives a local maximum f(1) = 2 − 9 + 12 + 15 = 20; f″(2) = 6 > 0, so x = 2 gives a local minimum f(2) = 16 − 36 + 24 + 15 = 19.

Notice that the local maximum (20) is only slightly above the local minimum (19), and f goes to +∞ as x → ∞. A local maximum need not be the largest value of the function.

Absolute maximum and minimum on a closed interval

A continuous function on a closed interval [a, b] always attains an absolute maximum and an absolute minimum. To find them, list the critical points inside (a, b), evaluate f at those points and at the end points a and b, and pick the largest and smallest values. No derivative test is needed.

Worked example: Find the absolute maximum and minimum of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].
Solution: f′(x) = 6x² − 30x + 36 = 6(x − 2)(x − 3), so the critical points are 2 and 3. f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 250 − 375 + 180 + 1 = 56. Absolute maximum 56 at x = 5; absolute minimum 24 at x = 1. Both occur at end points, which is exactly why end points must be checked.

Optimisation problems

Choose one variable, express the quantity to be maximised or minimised as a function of it using the given constraint, note the allowed range of the variable, and then apply the tests. Always state the answer in the units of the problem.

Worked example: An open box is made from a square tin sheet of side 18 cm by cutting equal squares of side x from each corner and folding up the sides. Find x for maximum volume.
Solution: V = x(18 − 2x)² with 0 < x < 9. Expanding, V = 4x³ − 72x² + 324x, so dV/dx = 12x² − 144x + 324 = 12(x − 3)(x − 9). In the allowed range, only x = 3 is critical. d²V/dx² = 24x − 144 = −72 < 0 at x = 3, so this is a maximum. Maximum volume = 3 × 12² = 432 cm³, with x = 3 cm.

Two results that save time: of all rectangles with a fixed perimeter, the square has the largest area; and for two positive numbers with a fixed sum, the product is largest when they are equal.

Tangents and normals

This topic was removed from the rationalised NCERT book, but it is still in the MHT‑CET and JEE Advanced syllabus. The slope of the tangent to y = f(x) at (x₀, y₀) is m = f′(x₀).

If f′(x₀) = 0 the tangent is horizontal and the normal is the vertical line x = x₀. For the curve y = x² at (1, 1), the tangent has slope 2 (so y = 2x − 1) and the normal has slope −1/2.

Common mistakes: (1) Substituting numerical values before differentiating in rate problems. (2) Treating f′(c) = 0 as proof of a maximum or minimum without a test. (3) Forgetting the end points when finding absolute extrema on a closed interval. (4) Reporting the value of x when the question asks for the maximum value of f, or the other way round. (5) Ignoring the physical range of the variable (for the box, x must lie between 0 and 9).

JEE and MHT‑CET focus

• Related rates: area, volume, shadow and ladder type problems, with correct units.
• Intervals of increase and decrease by sign analysis of f′, including functions with parameters (find k so that f is increasing on R).
• Local maxima and minima by both tests, and recognising points of inflection.
• Absolute extrema on closed intervals, including trigonometric functions.
• Optimisation of areas, volumes and costs, especially boxes, cylinders and cones.
• Equations of tangents and normals, and points where the tangent is parallel to an axis or a given line.

Practice questions