Application of Derivatives

The derivative of a function represents the instantaneous rate of change of one quantity with respect to another. If a quantity y is a function of x, that is, y = f(x), then the rate of change of y with respect to x at a point x = x₀ is given by the value of the derivative dy/dx at x = x₀, written as [dy/dx]_x=x₀.

Very often, two or more quantities change with respect to time, and they are connected to each other through some geometric or physical relation. Such problems are called rate of change or related-rates problems. The general method is: (i) write down the equation connecting the variables, (ii) differentiate both sides with respect to time t using the chain rule, and (iii) substitute the given numerical values only after differentiating.

Worked Example: The radius of a circle is increasing at a uniform rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

• Let r be the radius and A the area of the circle at time t, so A = πr².
• Differentiating with respect to t: dA/dt = 2πr · (dr/dt).
• Given dr/dt = 3 cm/s and r = 10 cm.
• dA/dt = 2π(10)(3) = 60π cm²/s.

So the area is increasing at the rate of 60π cm²/s (≈ 188.5 cm²/s) at that instant.

This same idea extends to problems involving volume, surface area, distance and speed, and other physically meaningful quantities — the derivative always tells us how fast one quantity changes as another changes.

A function f is said to be increasing on an interval (a, b) if x₁ < x₂ implies f(x₁) ≤ f(x₂) for all x₁, x₂ in (a, b), and strictly increasing if f(x₁) < f(x₂) whenever x₁ < x₂. Similarly, f is decreasing (or strictly decreasing) if the inequality reverses.

Using derivatives, if f is differentiable on (a, b):

• f is increasing on (a, b) if f′(x) ≥ 0 for every x in (a, b).
• f is decreasing on (a, b) if f′(x) ≤ 0 for every x in (a, b).
• f is strictly increasing if f′(x) > 0, and strictly decreasing if f′(x) < 0, throughout the interval (except possibly at isolated points).

Worked Example: Find the intervals in which f(x) = 2x³ − 15x² + 36x + 1 is increasing or decreasing.

• f′(x) = 6x² − 30x + 36 = 6(x² − 5x + 6) = 6(x − 2)(x − 3).
• f′(x) = 0 at x = 2 and x = 3; these split the real line into three intervals: (−∞, 2), (2, 3), (3, ∞).
• For x < 2 (say x = 0): f′(0) = 6(−2)(−3) = 36 > 0 ⟹ increasing.
• For 2 < x < 3 (say x = 2.5): f′(2.5) = 6(0.5)(−0.5) = −1.5 < 0 ⟹ decreasing.
• For x > 3 (say x = 4): f′(4) = 6(2)(1) = 12 > 0 ⟹ increasing.

So f is strictly increasing on (−∞, 2) ∪ (3, ∞) and strictly decreasing on (2, 3).

The derivative f′(x₀) gives the slope of the tangent to the curve y = f(x) at the point (x₀, y₀). The normal at that point is the line perpendicular to the tangent there.

If m = [dy/dx] at (x₁, y₁) is the slope of the tangent, then:

• Equation of tangent: y − y₁ = m(x − x₁).
• Slope of normal = −1/m (since tangent and normal are perpendicular).
• Equation of normal: y − y₁ = (−1/m)(x − x₁), provided m ≠ 0.
• If m = 0, the tangent is horizontal (y = y₁) and the normal is vertical (x = x₁). If the tangent itself is vertical, the normal is horizontal.

Worked Example: Find the equations of the tangent and normal to the curve y = x³ − x at the point (1, 0).

• dy/dx = 3x² − 1, so at x = 1: slope of tangent m = 3(1) − 1 = 2.
• Equation of tangent: y − 0 = 2(x − 1) ⟹ y = 2x − 2.
• Slope of normal = −1/m = −1/2.
• Equation of normal: y − 0 = (−1/2)(x − 1) ⟹ x + 2y − 1 = 0.

These ideas are also used to find points on a curve where the tangent is parallel or perpendicular to a given line, by equating f′(x) to the appropriate slope.

For a small change Δx in x, the corresponding change in y = f(x) can be approximated using the derivative, without recomputing f exactly at the new point. This is the method of differentials.

Let Δy = f(x + Δx) − f(x) be the actual change in y. The differential dy is defined as dy = f′(x) · dx, where dx = Δx is treated as a small increment. For small Δx, Δy ≈ dy, so:

f(x + Δx) ≈ f(x) + f′(x) · Δx

This lets us estimate the value of a function near a point where the exact value is easy to compute (like a perfect square or cube), without a calculator.

Worked Example: Use differentials to approximate √25.2.

• Let f(x) = √x. Choose x = 25 (a perfect square close to 25.2) and Δx = 0.2.
• f′(x) = 1/(2√x), so f′(25) = 1/(2×5) = 1/10 = 0.1.
• dy = f′(x)·Δx = 0.1 × 0.2 = 0.02.
• f(25.2) ≈ f(25) + dy = 5 + 0.02 = 5.02.

So √25.2 ≈ 5.02 (the true value is 5.01996…, confirming the approximation is accurate).

The same differential dy is also used to estimate the approximate error, or percentage error, in a computed quantity (such as volume or area) when there is a small error in a measured quantity (such as length or radius).

A point c is called a point of local maximum of f if f(c) ≥ f(x) for all x in some neighbourhood of c, and a point of local minimum if f(c) ≤ f(x) for all x near c. The value f(c) is then called a local maximum or local minimum value. Points where f′(x) = 0 (or f′ does not exist) are called critical points — every local maximum or minimum in the interior of the domain occurs at a critical point.

First Derivative Test: Let c be a critical point of f, that is, f′(c) = 0.

• If f′(x) changes sign from positive to negative as x increases through c, then c is a point of local maximum.
• If f′(x) changes sign from negative to positive as x increases through c, then c is a point of local minimum.
• If f′(x) does not change sign as x passes through c, then c is neither a local maximum nor a local minimum — it is called a point of inflection.

Worked Example: Find the local maxima and local minima of f(x) = x³ − 6x² + 9x + 15 using the first derivative test.

• f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
• Critical points: f′(x) = 0 ⟹ x = 1, x = 3.
• For x < 1 (say x = 0): f′(0) = 3(−1)(−3) = 9 > 0.
• For 1 < x < 3 (say x = 2): f′(2) = 3(1)(−1) = −3 < 0.
• For x > 3 (say x = 4): f′(4) = 3(3)(1) = 9 > 0.

Since f′ changes from + to − at x = 1, x = 1 is a point of local maximum, with local maximum value f(1) = 1 − 6 + 9 + 15 = 19. Since f′ changes from − to + at x = 3, x = 3 is a point of local minimum, with local minimum value f(3) = 27 − 54 + 27 + 15 = 15.

Second Derivative Test: Let f be twice differentiable at a critical point c, where f′(c) = 0.

• If f″(c) < 0, then c is a point of local maximum, and f(c) is the local maximum value.
• If f″(c) > 0, then c is a point of local minimum, and f(c) is the local minimum value.
• If f″(c) = 0, the test is inconclusive — fall back on the first derivative test.

Worked Example: Using the second derivative test, verify the nature of the critical points of f(x) = x³ − 6x² + 9x + 15 (critical points x = 1, 3, from the previous section).

• f″(x) = 6x − 12.
• At x = 1: f″(1) = 6 − 12 = −6 < 0 ⟹ local maximum (confirms the earlier result).
• At x = 3: f″(3) = 18 − 12 = 6 > 0 ⟹ local minimum (confirms the earlier result).

To find the absolute (global) maximum and minimum of a continuous function f on a closed interval [a, b], we use a simple procedure: find all critical points of f inside (a, b), then evaluate f at each of these critical points and at the two endpoints a and b. The largest of all these values is the absolute maximum, and the smallest is the absolute minimum.

Worked Example: Find the absolute maximum and absolute minimum of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].

• f′(x) = 6x² − 30x + 36 = 6(x − 2)(x − 3) ⟹ critical points x = 2, 3, both lying in (1, 5).
• f(1) = 2 − 15 + 36 + 1 = 24.
• f(2) = 16 − 60 + 72 + 1 = 29.
• f(3) = 54 − 135 + 108 + 1 = 28.
• f(5) = 250 − 375 + 180 + 1 = 56.

Comparing all four values {24, 29, 28, 56}, the absolute maximum is 56 at x = 5, and the absolute minimum is 24 at x = 1. Note that neither the absolute maximum nor the absolute minimum occurred at the local extrema here — this is why endpoints must always be checked on a closed interval.