The derivative of a function tells you its rate of change — but in NDA, you're almost never asked to compute a derivative for its own sake. You're asked what that derivative implies about the function's behavior: is it climbing or falling, and where does it turn around?
Here's the core idea. Imagine driving on a hill road. When you're going uphill, your altitude is increasing — the slope is positive. When you crest the hill and start coming down, the slope briefly hits zero at the peak, then goes negative. That instant at the top — where slope = 0 — is a critical point, and the altitude at that instant is a local maximum.
Formally: if f'(x) > 0 on an interval, the function is increasing there. If f'(x) < 0, it's decreasing. The points where f'(x) = 0 (or where f'(x) doesn't exist but f does) are critical points — candidates for local maxima or minima.
A local maximum at x = c means f(c) ≥ f(x) for all x near c. A global (absolute) maximum means f(c) ≥ f(x) for all x in the domain — and here's a trap NDA loves: on an open interval, a function can approach but never attain its supremum. The function f(x) = x on (-1, 1) increases throughout; its values approach 1 and -1 but never reach them. No global max, no global min — even though the function is perfectly well-behaved.
Think of monotonicity like a one-way street: an increasing function never doubles back, a decreasing function never climbs. Knowing where the "street direction changes" (at critical points) gives you the full picture of the function's shape.
The practical payoff: NDA questions on this topic reduce to three operations — differentiate, factor/sign-analyze, and classify. If you can do all three fast, these are free marks.
Set f'(x) = 0 and solve. Also check where f'(x) is undefined (but f(x) is defined) — those are also critical points.
For f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7:
f'(x) = x^2 - 5x + 6 = (x-2)(x-3)
Critical points: x = 2 and x = 3.
Draw a number line, place the critical points, and test one value in each interval:
| Interval | Test point | f'(x) sign | Function |
|---|---|---|---|
| x < 2 | x = 0 | (−2)(−3) = + | Increasing |
| 2 < x < 3 | x = 2.5 | (0.5)(−0.5) = − | Decreasing |
| x > 3 | x = 4 | (2)(1) = + | Increasing |
So f increases on (−∞, 2], decreases on [2, 3], and increases on [3, +∞). Note the closed brackets — a function is said to be increasing/decreasing on a closed interval when f' ≥ 0 (or f' ≤ 0) throughout, with equality allowed at the endpoints.
First Derivative Test: At x = c, if f' changes from + to −, it's a local maximum. If from − to +, it's a local minimum. If no sign change, it's a point of inflection (no extremum).
From the table above: at x = 2, f' goes from + to − — local maximum. At x = 3, f' goes from − to + — local minimum.
Second Derivative Test: Faster when the algebra cooperates. At a critical point c:
f''(c) > 0 — local minimum (curve is concave up, like a bowl)f''(c) < 0 — local maximum (curve is concave down, like a hill)f''(c) = 0 — test is inconclusive; use the first derivative testFor f(x) = x / \ln x (with x > 1): differentiate once to get f'(x) = \frac{\ln x - 1}{(\ln x)^2}. At x = e: \ln e = 1, so f'(e) = 0 — critical point confirmed. Differentiate again using quotient rule:
f''(x) = \frac{2 - \ln x}{x(\ln x)^3}
At x = e: f''(e) = \frac{2 - 1}{e \cdot 1} = \frac{1}{e} > 0 — local minimum. The minimum value is f(e) = \frac{e}{\ln e} = \frac{e}{1} = e.
This is where NDA specifically tests your precision. On an open interval (a, b):
(a, b).f(x) = x on (-1, 1): strictly increasing, supremum is 1 — but 1 is never reached. No maximum, no minimum. Full stop.On a closed interval [a, b], a continuous function always attains both its maximum and minimum (Extreme Value Theorem). That's why the interval type in the question is the first thing you should look at.
For f(x) = 2\cos^2 x - 1: recognize this as \cos(2x). Then f'(x) = -2\sin(2x).
For decreasing: f'(x) < 0 \Rightarrow \sin(2x) > 0 \Rightarrow 2x \in (0, \pi) \Rightarrow x \in (0, \pi/2).
Length of this interval: \pi/2. This is the longest single interval where f decreases.
For a function to have no local maximum or minimum on x > 0, its derivative must not change sign for x > 0. This means f'(x) = 0 has no positive real roots.
For f(x) = \frac{16x^3}{3} - 4bx^2 + x: f'(x) = 16x^2 - 8bx + 1. This is a quadratic in x. For no real roots: discriminant < 0.
\Delta = (8b)^2 - 4 \cdot 16 \cdot 1 = 64b^2 - 64 = 64(b^2 - 1)
\Delta < 0 \Rightarrow b^2 < 1 \Rightarrow -1 < b < 1
Since b is non-negative: 0 \leq b < 1. At b = 0, f'(x) = 16x^2 + 1 > 0 always — confirmed no extrema.
When f'(x) is a product of linear factors like (x - a)(x - b) with a < b, you don't need to test points. The sign pattern is always: +, −, + (positive outside, negative between). For three factors (x-a)(x-b)(x-c) with a < b < c, it's −, +, −, +. Just count from the right: rightmost interval is always positive for even-degree leading coefficient. Standard number-line method: 4 steps. This pattern: 1 step. Saves 20-30 seconds per question.
Before differentiating any trig function in a monotonicity question, check if it simplifies. 2\cos^2 x - 1 = \cos(2x). 1 - 2\sin^2 x = \cos(2x). 2\sin x \cos x = \sin(2x). Differentiating the simplified form takes 5 seconds; differentiating the expanded form and then simplifying takes 40 seconds. Commit these three double-angle collapses to reflex memory.
Before attempting any max/min question, read the interval notation. Parentheses (a, b) = open = endpoints excluded. Brackets [a, b] = closed = endpoints included. If the question asks "does the function attain its maximum on (-1, 1)?" and the function is monotone on that interval, the answer is immediately "No" — no computation needed. This eliminates wrong answer choices in under 5 seconds versus 60 seconds of blind calculation.
At a critical point, f''(c) > 0 means the curve is bending upward (concave up) — like a bowl — so you're at the bottom: minimum. f''(c) < 0 means bending downward — like a hill — so you're at the top: maximum. The mnemonic: positive second derivative points the curve up toward a minimum. If you confuse these, you'll flip every classification. Lock in the bowl image: bowl holds water, water sits at the bottom — that's your minimum, that's where f'' > 0.
When asked for values of a parameter that ensure a function has no local extrema, compute f'(x) and check whether the resulting quadratic (or other equation) has real roots. No real roots = no sign change = no extremum. Condition: discriminant \Delta < 0. This converts an extremum-existence question into a pure discriminant inequality — 3 algebraic steps versus analyzing sign charts for each parameter value. Speed gain: roughly 4 steps down to 2.
In the exam hall, work through this decision tree in order:
1. Read the interval type first. Open interval + monotone function = no global extremum attained. If the question is purely about that, you're done.
2. Compute f'(x) and simplify. For trig functions, always apply identities before differentiating. Factor the derivative completely.
3. Find critical points. Set f'(x) = 0. Note any points where f' is undefined.
4. Sign analysis. Use the sign-sandwich pattern for factored polynomials. For transcendental functions, test one point per interval.
5. Classify. Use the second derivative test if f'' is easy to compute. Use the first derivative test (sign change) when f'' is messy or zero.
6. Answer the exact question asked. "Where does f decrease?" wants the interval. "What is S?" wants the closed interval. "Does f attain its minimum?" wants a yes/no based on interval type and behavior.
Never jump to the second derivative test reflexively — if f''(c) = 0, you've wasted time and still need the first derivative test. Save the second derivative test for when f'' evaluates cleanly (like f(x) = x/\ln x at x = e).
Why this question: This is the most commonly mishandled question type — NDA tests whether you confuse "function approaches a value" with "function attains a value". The open interval is the trap.
Solving path: f(x) = x on (-1, 1) is strictly increasing. f'(x) = 1 > 0 everywhere on the interval — no critical points. On the open interval, as x \to 1^-, f(x) \to 1, but x = 1 is not in the domain. The function never reaches 1. Similarly never reaches -1. Neither maximum nor minimum is attained. Answer: D.
Why this question: Tests whether you can quickly identify a trig identity and then set up a monotonicity condition — both steps must be reflexive to solve in under 90 seconds.
Solving path: Write f(x) = 2\cos^2 x - 1 = \cos(2x) immediately. Differentiate: f'(x) = -2\sin(2x). For decreasing: f'(x) < 0 \Rightarrow \sin(2x) > 0 \Rightarrow 2x \in (0, \pi) \Rightarrow x \in (0, \pi/2). Interval length = \pi/2 - 0 = \pi/2. Answer: C.
Why this question: This pair of questions (S and T) tests the closed-vs-open interval convention for monotonicity and the sign-sandwich pattern on a quadratic derivative.
Solving path: f'(x) = x^2 - 5x + 6 = (x-2)(x-3). Sign pattern: + for x < 2, − for 2 < x < 3, + for x > 3. For T (increasing): f'(x) \geq 0, which holds on x \leq 2 or x \geq 3 — closed intervals because the derivative is zero at the endpoints but non-negative. Answer: A.
Solving path: Same derivative. For S (decreasing): f'(x) \leq 0 when 2 \leq x \leq 3. The closed bracket [2, 3] is correct — at x = 2 and x = 3, f' = 0, which is consistent with a (weakly) decreasing function. Answer: D.
Why this question: Parameter-based extremum question — NDA uses these to check whether you know the discriminant shortcut for "no real roots = no extremum".
Solving path: f'(x) = 16x^2 - 8bx + 1. For no extremum in x > 0, this must have no positive real roots. Since the leading coefficient is positive (16 > 0), f'(x) > 0 always when discriminant < 0. \Delta = 64b^2 - 64 < 0 \Rightarrow b^2 < 1. With b \geq 0: 0 \leq b < 1. Answer: D.
Why this question: A three-statement question on second derivative test and minimum value — tests all three parts of the classification process in one shot.
Solving path: f(x) = x/\ln x, f'(x) = (\ln x - 1)/(\ln x)^2. At x = e: \ln e = 1, so f'(e) = 0. For f''(e): compute f''(x) = (2 - \ln x) / [x(\ln x)^3]. At x = e: f''(e) = (2-1)/(e \cdot 1) = 1/e > 0. Statement 1: true. Since f''(e) > 0, local minimum at x = e. Statement 2: true. f(e) = e/\ln e = e/1 = e. Statement 3: true. Answer: D (all three correct).
Attainment on open intervals. Writing that a monotone function on (-1, 1) has maximum f(0.9999...) is wrong — the supremum must be attained at a point in the domain. On an open interval with no critical points, there is simply no maximum or minimum, full stop.
Using closed intervals for monotonicity with strict inequalities. If f'(x) > 0 strictly on (a, b) and equals zero only at isolated endpoints, the function still increases on [a, b] — the \geq 0 condition (not strict > 0) is what matters for a closed monotonicity interval. Many students write open brackets when the question expects closed.
Forgetting to factor f'(x) completely. If you set x^2 - 5x + 6 = 0 numerically without factoring, you'll get critical points but miss the sign pattern instantly. Always factor before doing sign analysis.
Second derivative test on f''(c) = 0. If you compute f''(c) = 0, the test gives you no information. You must fall back to the first derivative test (sign change of f'). Reporting "inconclusive" and stopping is a mistake — go back and check the sign change.
Confusing the function y = 2f(x) + ax - b with f(x). In questions where a new function y is formed from f, you need y'(0) = 0 as the condition for an extremum at x = 0 — not f'(0) = 0. Here y'(0) = 2f'(0) + a = a. The vertical shift b is irrelevant to the location of extrema entirely.
Applying the trig monotonicity interval without checking for period. \sin(2x) > 0 holds on (0, \pi/2), but also on (\pi, 3\pi/2), etc. The longest single continuous interval of decrease is \pi/2, not the full union. Read whether the question asks for one interval's length or the total set.