If you can find a slope, you can solve half of NDA calculus. Derivatives measure how fast one quantity changes with respect to another — the slope of a curve, the speed of a body, the rate a balloon inflates. In this Cavalier lesson you will learn the standard rules, then put them to work on tangents, normals, rates of change, and maxima & minima, the four application types the NDA loves.
Why Derivatives Carry So Many Marks
Derivatives appear in 6 to 10 questions in a typical NDA Maths paper, either directly or hidden inside physics-style word problems. The good news: almost every question reduces to one of four standard tasks, so once you drill the rules you can score these marks fast and move on.
A derivative answers a simple question: if the input changes a tiny bit, how much does the output change? That ratio of change is the derivative. Geometrically it is the slope of the tangent line; physically it is a rate. The same number describes the steepness of a graph and the speed of a moving particle — which is why this one idea connects geometry, algebra and physics in the NDA syllabus.
Because the exam is purely objective with negative marking, your aim is accuracy with speed. Differentiation rewards exactly that: once the rules are automatic, you can compute a slope or classify a turning point in under a minute, leaving more time for the trickier coordinate-geometry and probability questions.
The derivative of f(x) is written f′(x) or dy÷dx. It gives the instantaneous rate of change of y with respect to x — not the average over an interval.
The Derivative from First Principle
The formal definition every NDA aspirant should be able to write is the limit of the difference quotient:
f′(x) = limh→0 [ f(x+h) − f(x) ] ÷ h
This single formula is the source of every shortcut rule you will memorise next.
You rarely use it in the exam directly, but the NDA does occasionally ask you to differentiate a simple function “from first principle”, so know the structure: build f(x+h), subtract f(x), divide by h, then let h shrink to zero. The idea is that as h becomes smaller and smaller, the secant line joining two nearby points on the curve gets closer and closer to the tangent at a single point — the difference quotient turns into the exact slope.
For instance, differentiating f(x) = x² from first principle gives [ (x+h)² − x² ] ÷ h = (2xh + h²) ÷ h = 2x + h, and as h → 0 this becomes 2x. That matches the power rule — which is exactly why the shortcut rules are trustworthy: each one is a packaged result of this same limit.
If a question says “by first principle”, you must show the limit. Using the power rule directly will not earn the answer they intend.
Standard Derivatives You Must Memorise
These are the building blocks. Burn them into memory — the exam never gives you time to derive them.
- d÷dx (xn) = n·xn−1
- d÷dx (sin x) = cos x
- d÷dx (cos x) = −sin x
- d÷dx (tan x) = sec2 x
- d÷dx (ex) = ex
- d÷dx (ln x) = 1÷x
- d÷dx (ax) = ax·ln a
- d÷dx (constant) = 0
The derivative of any constant is 0, and the derivative of x is 1. These trivial facts trip up students under time pressure.
The Four Rules That Combine Functions
Real questions mix functions together. You combine standard derivatives using four rules.
Sum and difference
(u ± v)′ = u′ ± v′. Differentiate term by term.
Product rule
(u·v)′ = u′v + uv′
Quotient rule
(u÷v)′ = (u′v − uv′) ÷ v2
Chain rule
For a function inside a function, y = f(g(x)):
dy÷dx = f′(g(x)) · g′(x)
Differentiate the outer layer, then multiply by the derivative of the inner layer.
Forgetting the chain rule’s inner factor. d÷dx (sin 2x) is 2 cos 2x, not cos 2x. The extra ×2 comes from differentiating the inside (2x).
Tangents and Normals to a Curve
The slope of the tangent to y = f(x) at a point is just the derivative evaluated there: m = f′(x1). The normal is perpendicular to the tangent, so its slope is the negative reciprocal.
Tangent slope: m = f′(x1)
Normal slope: −1÷m
Tangent line: y − y1 = m(x − x1)
Method: differentiate, substitute the point’s x-value to get m, then plug into the point-slope line equation. For the normal, replace m with −1÷m and use the same point. Both lines pass through the point of contact, so the only thing that changes between them is the slope.
A handy self-check: the tangent and normal at any point are perpendicular, so the product of their slopes must equal −1. If your two slopes do not multiply to −1, you have made an arithmetic slip somewhere and should redo the step.
A horizontal tangent means f′(x) = 0; a vertical tangent means f′(x) is undefined (slope → ∞). Watch for questions asking where the tangent is parallel to the x-axis.
Rate of Change and Related Rates
When a quantity changes with time, its derivative with respect to time is its rate. If a balloon’s radius grows, dV÷dt and dr÷dt are linked through the chain rule — these are related-rate problems.
Velocity = ds÷dt (rate of change of position). Acceleration = dv÷dt = d2s÷dt2 (second derivative of position).
Strategy for related rates:
- Write the equation linking the two quantities (e.g. V = (4÷3)πr3).
- Differentiate both sides with respect to time t.
- Substitute the known instantaneous values and solve for the unknown rate.
Plugging in the given numbers before differentiating. Differentiate the general relation first, then substitute — otherwise constants vanish and you lose the link between the rates.
Increasing and Decreasing Functions
The sign of the first derivative tells you which way a function is heading.
If f′(x) > 0 on an interval → f is increasing there.
If f′(x) < 0 on an interval → f is decreasing there.
If f′(x) = 0 → a stationary (critical) point.
To find where a function increases or decreases: compute f′(x), set it to zero to find critical points, and test the sign of f′ in each interval between them. Pick any convenient test value inside an interval, evaluate f′ there, and the sign you get holds for the whole interval — you do not need to check every point one by one.
This sign analysis is also the backbone of curve sketching and of proving simple inequalities, both of which occasionally surface in NDA Maths. Knowing where a function rises and where it falls tells you the overall shape of its graph without plotting dozens of points.
A function is monotonically increasing if f′(x) ≥ 0 for all x in the domain. The NDA often phrases options this way — the ≥ (not strict >) is deliberate.
Maxima and Minima
This is the highest-yield application. A local maximum or minimum occurs at a critical point where f′(x) = 0. To classify it, use the second derivative test.
At a critical point c where f′(c) = 0:
If f″(c) < 0 → local maximum
If f″(c) > 0 → local minimum
If f″(c) = 0 → test fails; use the first-derivative sign change.
Full procedure:
- Differentiate to get f′(x).
- Solve f′(x) = 0 for critical points.
- Find f″(x) and check its sign at each critical point.
- Substitute back into f(x) to get the actual maximum or minimum value.
Stopping at the x-value of the critical point. If the question asks for the maximum value, you must substitute x back into f(x), not report x itself.
A point of inflection is where f″(x) = 0 and the curve changes concavity. It is not a max or min — the slope does not reverse sign.
Worked Example: Maxima and Minima
Find the local maximum and minimum values of f(x) = x3 − 3x + 2.
So the local maximum value is 4 (at x = −1) and the local minimum value is 0 (at x = +1).
Notice how clean the work stays when you follow the four steps in order. That discipline is what saves time in the real exam. The cubic gave two critical points, and the second derivative immediately separated the peak from the valley without any sign tables — this is why the second-derivative test is the fastest route for polynomial questions.
Worked Example: Tangent and Rate
Find the slope of the tangent to y = x² + 3x at the point where x = 2, and state how fast y changes there as x increases.
The slope is 7, meaning y increases 7 units for each unit increase in x at that point. The normal there would have slope −1÷7.
The phrase “how fast y changes with x” is just asking for dy÷dx. Treat “rate” and “slope” as the same number when the variable is x.
Previous-Year Style Question
Q. The function f(x) = x3 − 6x2 + 9x + 5 has a local maximum at which value of x?
Answer: f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Setting f′(x) = 0 gives x = 1 and x = 3. Now f″(x) = 6x − 12. At x = 1, f″ = −6 < 0 → local maximum. At x = 3, f″ = +6 > 0 → local minimum. So the local maximum is at x = 1.
This is the exact template the NDA reuses: a cubic, factorise the derivative, apply the second-derivative test. Practise it until the steps feel automatic. The only variation examiners add is sometimes asking for the maximum value f(1) instead of the location x = 1, or swapping the cubic for a product of functions that needs the product rule first — the core method never changes.
Quick Revision
- Derivative = instantaneous rate of change = slope of tangent = f′(x).
- Power rule, product, quotient, chain — the four engines of differentiation.
- Tangent slope = f′(x1); normal slope = −1÷f′(x1).
- f′ > 0 increasing, f′ < 0 decreasing, f′ = 0 critical point.
- Second-derivative test: f″ < 0 max, f″ > 0 min.
- Always substitute x back to get the actual maximum or minimum value.
- For related rates, differentiate the general relation first, then plug in numbers.
Drill ten cubic max-min problems and ten tangent problems before your next mock — these two types alone can secure most of the calculus marks in NDA Maths.
Frequently asked questions
How many questions on derivatives appear in NDA Maths?
Derivatives and their applications typically account for 6 to 10 questions per paper, including word problems on rates and maxima-minima. It is one of the highest-weight calculus topics, so mastering it gives a strong scoring return.
What is the difference between a tangent and a normal?
The tangent touches the curve at a point and has slope f′(x). The normal is perpendicular to the tangent at the same point, so its slope is the negative reciprocal, −1÷f′(x).
When does the second-derivative test fail?
It fails when f″(c) = 0 at a critical point, giving no conclusion. In that case you fall back on the first-derivative test: check whether f′ changes sign from positive to negative (maximum) or negative to positive (minimum) around c.
What is a point of inflection?
A point of inflection is where the curve changes its concavity, occurring where f″(x) = 0 and the second derivative changes sign. It is not a maximum or minimum because the slope does not reverse direction there.
Why must I differentiate before substituting in related-rate problems?
If you substitute the given instantaneous values first, the variables become constants and their derivatives vanish, breaking the link between the rates. Always differentiate the general relation with respect to time, then plug in the numbers.
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