Application of Derivatives Class 12 Formulas

This page collects the formulas you need from Application of Derivatives (NCERT Class 12 Mathematics, Part I, Chapter 6): rate of change of quantities, the chain rule for related rates, increasing and decreasing functions, critical points, the first and second derivative tests for local maxima and minima, and the working rule for absolute maxima and minima on a closed interval.

Each formula is grouped by topic, with the meaning of every symbol, a line on when to use it, and worked examples with original numbers. For the explanations and derivations of these tests (including the Mean Value Theorem proof of Theorem 1), see the detailed Class 12 Maths Formulas hub and the chapter notes.

Formulas at a Glance

All Formulas, Grouped by Topic

Rate of Change of Quantities

If one quantity \( y \) varies with another quantity \( x \) by the rule \( y = f(x) \), then the derivative \( \frac{dy}{dx} \) (or \( f'(x) \)) is the rate of change of \( y \) with respect to \( x \) (NCERT, p. 148):

\[ \frac{dy}{dx} = f'(x) \]

At a particular value \( x = x_0 \), this rate is written as follows:

\[ \left. \frac{dy}{dx} \right|_{x=x_0} = f'(x_0) \]

If \( x \) and \( y \) both vary with time \( t \), the rate of change of \( y \) with respect to \( x \) is obtained by the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{dt} \big/ \frac{dx}{dt}, \quad \text{provided } \frac{dx}{dt} \neq 0 \]

This is the formula behind nearly every “how fast is it changing” problem: write \( A = \pi r^2 \), differentiate with respect to \( t \), and get \( \frac{dA}{dt} = 2\pi r \, \frac{dr}{dt} \).

Increasing and Decreasing Functions

A real-valued function \( f \) is strictly increasing on an interval if bigger \( x \) always gives a bigger \( f(x) \), and strictly decreasing if bigger \( x \) always gives a smaller \( f(x) \) (NCERT, p. 153). The derivative gives a working test (NCERT, p. 154): let \( f \) be continuous on \( [a, b] \) and differentiable on \( (a, b) \). Then

• \( f \) is increasing in \( [a, b] \) if \( f'(x) \ge 0 \) for each \( x \in (a, b) \);
• \( f \) is decreasing in \( [a, b] \) if \( f'(x) \leq 0 \) for each \( x \in (a, b) \);
• \( f \) is a constant function in \( [a, b] \) if \( f'(x) = 0 \) for each \( x \in (a, b) \).

To find the intervals, solve \( f'(x) = 0 \): the roots split the real line (or the given domain) into disjoint intervals, and the sign of \( f'(x) \) in each interval tells you whether \( f \) is increasing there (NCERT, p. 155).

Critical Points, Local Maxima and Minima

A point \( c \) in the domain of \( f \) is a critical point if either \( f'(c) = 0 \) or \( f \) is not differentiable at \( c \) (NCERT, p. 164). If \( f \) has a local maximum or local minimum at \( c \), then \( c \) must be a critical point — but the converse is false: \( f'(c) = 0 \) does not guarantee an extremum.

For \( f(x) = x^3 \), \( f'(0) = 0 \) yet 0 is neither a local maximum nor a local minimum (NCERT, p. 164).

First Derivative Test

Let \( f \) be continuous at a critical point \( c \). Examine the sign of \( f'(x) \) as \( x \) increases through \( c \) (NCERT, p. 164):

• If \( f'(x) \) changes sign from positive to negative through \( c \), then \( c \) is a point of local maxima.
• If \( f'(x) \) changes sign from negative to positive through \( c \), then \( c \) is a point of local minima.
• If \( f'(x) \) does not change sign through \( c \), then \( c \) is neither — it is a point of inflexion.

Second Derivative Test

Let \( f \) be twice differentiable at \( c \) and let \( f'(c) = 0 \) (NCERT, p. 166):

• \( x = c \) is a point of local maxima if \( f”(c) \lt 0 \); \( f(c) \) is the local maximum value.
• \( x = c \) is a point of local minima if \( f”(c) \gt 0 \); \( f(c) \) is the local minimum value.
• The test fails if \( f”(c) = 0 \) — go back to the first derivative test.

Why the sign of \( f” \) decides: at a local maximum the slope falls from positive to negative, so the slope itself is decreasing, which means \( f” \lt 0 \); at a local minimum the slope increases from negative to positive, so \( f” \gt 0 \).

Absolute Maxima and Minima on a Closed Interval

Every continuous function on a closed interval \( [a, b] \) attains an absolute maximum and an absolute minimum (NCERT, p. 171). The working rule (NCERT, p. 172):

1. Find all critical points of \( f \) in \( [a, b] \) — points where \( f'(x) = 0 \) or \( f \) is not differentiable.
2. Take the end points \( a \) and \( b \).
3. Calculate \( f \) at every point from Step 1 and Step 2.
4. The largest of these values is the absolute maximum; the smallest is the absolute minimum.

If an absolute maximum or minimum occurs at an interior point \( c \), then \( f'(c) = 0 \) (NCERT, p. 172).

What Each Symbol Means

When to Use Each Formula

A quick way to remember the sign rules: positive derivative → function climbs; negative derivative → function falls. For the second derivative, think “above zero is a valley”: \( f”(c) \gt 0 \) ⇒ local minimum; \( f”(c) \lt 0 \) ⇒ local maximum.

Worked Examples

Example 1: Rate of change — area of a circle

Example 2: Using the chain rule — a balloon’s volume

Example 3: Second derivative test — local extrema

Practise applying these formulas on the textbook’s own questions: Exercise 6.1 (rates), Exercise 6.2 (increasing/decreasing), Exercise 6.3 (maxima and minima) and the Miscellaneous Exercise.

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between a local maximum and an absolute maximum?

A local maximum is the largest value of \( f \) in a small neighbourhood around the point \( c \); the absolute (global) maximum is the largest value of \( f \) on the whole given interval.

On a closed interval, the absolute maximum can occur at an end point, but a local maximum requires an interior point \( c \) where \( f(c) \geq f(x) \) for all \( x \) in \((c-h, c+h) \) (NCERT, p. 163). The graph in Fig 6.19 shows both cases.

When does the second derivative test fail?

It fails when \( f'(c) = 0 \) and \( f”(c) = 0 \) at the same point. In that situation the test gives no information, so you must go back to the first derivative test and check whether \( f'(x) \) changes sign through \( c \) (NCERT, p. 166). The derivative may change sign (extremum) or keep its sign (inflexion).

How do I find the intervals in which a function is increasing or decreasing?

Find \( f'(x) \), solve \( f'(x) = 0 \), and let the roots divide the domain into intervals. On each interval, test the sign of \( f'(x) \) using one convenient value of \( x \). Where \( f'(x) \gt 0 \), the function is increasing; where \( f'(x) \lt 0 \), it is decreasing (NCERT, p. 155).

Why is \( f'(c) = 0 \) not enough to guarantee an extremum?

Because the slope can be zero without the graph turning. For \( f(x) = x^3 \), \( f'(0) = 3(0)^2 = 0 \), but the graph keeps rising through \( x = 0 \) — it is a point of inflexion, not a turning point (NCERT, p. 164). You must apply a test that examines the sign of \( f'(x) \) or the sign of \( f”(x) \).

Reference: NCERT Class 12 Mathematics textbook, chapter Application of Derivatives. For the full chapter, see the official NCERT Class 12 Maths textbook (Part I) and the Maths formulas index for the whole syllabus.