Differentiation is the process of finding the rate at which one quantity changes with respect to another. If you have a function f(x), its derivative f'(x) tells you the instantaneous slope of the curve at any point — how steeply the function is rising or falling right there.
Here is a concrete way to think about it. Imagine you are driving from Delhi to Agra. Your average speed over the whole journey is total distance divided by total time. But your speedometer shows something different — it shows your speed at this exact moment, right now. That is the derivative. Mathematically, it is the limit of the average rate of change as the time interval shrinks to zero:
For NDA, you will almost never compute this limit directly. The exam tests whether you can apply the standard rules — and whether you spot the traps (modulus functions, piecewise definitions, corners in graphs) where derivatives fail to exist.
The notation you will see interchangeably: f'(x), dy/dx, D[y], ẏ. They all mean the same thing. Get comfortable with all four — NDA questions freely switch between them.
One more thing worth fixing in your head: differentiation is a linear operator. That means:
This linearity is why you can break apart sums and pull out constants without any ceremony. Where differentiation gets interesting — and where NDA sets traps — is products, quotients, compositions, and functions defined via absolute values or piecewise conditions.
These are not optional memorization — they are the alphabet of differentiation:
| Function | Derivative |
|---|---|
| xⁿ | nxⁿ⁻¹ |
| eˣ | eˣ |
| aˣ | aˣ ln a |
| ln x | 1/x |
| sin x | cos x |
| cos x | -sin x |
| tan x | sec²x |
| cot x | -cosec²x |
| sec x | sec x · tan x |
| cosec x | -cosec x · cot x |
| sin⁻¹ x | 1/√(1-x²) |
| cos⁻¹ x | -1/√(1-x²) |
| tan⁻¹ x | 1/(1+x²) |
Notice the pattern for inverse trig: sin⁻¹ and cos⁻¹ are negatives of each other, and tan⁻¹ has 1+x² in the denominator instead of √(1-x²). NDA loves testing whether you mix these up.
Use the chain rule whenever you have a function inside a function — a composite function.
In plain language: differentiate the outer function (leaving the inner function untouched), then multiply by the derivative of the inner function.
Example: d/dx[sin(3x + 2)]
sin(u) → derivative is cos(u)u = 3x + 2 → derivative is 3cos(3x + 2) × 3 = 3cos(3x + 2)The chain rule extends to any number of compositions. For d/dx[sin²(3x)], you apply it twice: outer is the square, inner is sin(3x), which itself needs the chain rule.
When two differentiable functions are multiplied together, you cannot just multiply their derivatives. The product rule is:
Mnemonic for the exam hall: "derivative of first times second, plus first times derivative of second." Apply it left-to-right every time so you do not miss a term.
Example: y = x³eˣ
u = x³, u' = 3x²v = eˣ, v' = eˣdy/dx = 3x²eˣ + x³eˣBoth option A and option D in NDA questions often list this in different factored forms — 3x²eˣ + x³eˣ versus x²eˣ(3+x). They are the same. Do not get confused by factoring.
The order matters — it is always (derivative of numerator × denominator) minus (numerator × derivative of denominator), all over denominator squared. Swapping the subtraction order is one of the most common NDA errors.
Example: y = (x+1)/(x-1)
u = x+1, u' = 1v = x-1, v' = 1dy/dx = [1·(x-1) - (x+1)·1] / (x-1)²= [x - 1 - x - 1] / (x-1)² = -2/(x-1)²When y is not explicitly solved for — for example x² + y² = r² — differentiate both sides with respect to x, treating y as a function of x. Every time you differentiate a term involving y, multiply by dy/dx.
Example: Differentiate x² + y² = r²
2x + 2y(dy/dx) = 0dy/dx = -x/yThis is also how you derive d/dx[sin⁻¹(x)] = 1/√(1-x²). Let y = sin⁻¹(x), so sin y = x. Differentiate: cos y · (dy/dx) = 1, giving dy/dx = 1/cos y = 1/√(1-sin²y) = 1/√(1-x²).
Here is where NDA sets its most reliable traps. A function is differentiable at a point only if the left-hand derivative equals the right-hand derivative at that point.
For f(x) = |x|:
x > 0: f(x) = x, so f'(x) = 1x < 0: f(x) = -x, so f'(x) = -1x = 0: LHD = -1, RHD = 1 — they differ, so f'(0) does not exist.For f(x) = x/|x| when x < 0: since |x| = -x for negative x, we get x/(-x) = -1. The derivative of the constant -1 is 0.
The rule: whenever you see |x| or an absolute value expression, rewrite using the piecewise definition before attempting to differentiate, and always check whether you are being asked about a specific region or about a specific point.
Write the quotient rule as (Lo D-Hi minus Hi D-Lo) over Lo-Lo, where Hi = numerator, Lo = denominator, D = derivative. This vocal pattern forces you to always start with the denominator in the first term's coefficient position, preventing the single most common sign error. Standard written approach: 30s to recall and set up. With QUOD locked in: 10s. No more sign flip on the subtraction.
For any composite function, ask: "What is the last operation applied to x?" That outermost operation is what you differentiate first, leaving everything inside unchanged, then multiply by the derivative of everything inside. Example: d/dx[e^(sin x)] — last operation is the exponential, so: e^(sin x) × cos x. Standard approach (trying to work inside-out): high error rate, ~40s. Outside-in with peeling: ~12s, near-zero error once the habit is set.
The moment you see |something|, do not differentiate immediately. Rewrite: |g(x)| = g(x) if g(x) > 0, and |g(x)| = -g(x) if g(x) < 0. Then the question tells you which region you are in, and differentiation becomes trivial. This collapses a 3-step analysis into 1 step. On x/|x| for x < 0: rewrite as x/(-x) = -1, derivative = 0. Done in 8 seconds. Trying to apply a derivative formula directly: either wrong answer or 45 seconds of confusion.
sin⁻¹ and cos⁻¹ are a pair: their derivatives are identical except one is negative. d/dx[sin⁻¹x] = +1/√(1-x²), d/dx[cos⁻¹x] = -1/√(1-x²). Their sum is a constant, so their derivatives must cancel — that is why one is positive and one is negative. This logical anchor prevents the mix-up that NDA answer choices deliberately exploit. Recalling by logic: 5 seconds vs. brute memorization that fails under pressure.
Before reaching for the quotient rule on u/v, check if u/v can be rewritten as a product: u · v⁻¹. Sometimes differentiating v⁻¹ = (x-1)⁻¹ with the chain rule and multiplying is faster than setting up the full quotient-rule fraction. This saves one algebraic step on simple rational functions. For y = (x+1)(x-1)⁻¹: use quotient rule (2 terms to set up). For y = x·(x²+1)⁻¹: product rule with chain rule is 1 line. Choose based on which denominator derivative is simpler. Step saving: 2 lines of algebra on the right cases.
In the exam hall, run through this sequence the moment you see a differentiation question:
Step 1 — Identify the function type. Is it a product, a quotient, a composition, or a single standard form? More than one rule may apply — decide the outermost structure first.
Step 2 — Check for absolute values or piecewise definitions. If present, rewrite the function in the given domain before anything else.
Step 3 — Apply rules in order:
u'v + uv'Step 4 — Verify differentiability if the question asks for a value at a specific point. Check LHD = RHD if the function has a corner, absolute value, or piecewise structure near that point.
Step 5 — Cross-check against the options. NDA options often include factored and expanded forms of the same answer. If your answer matches one option after factoring or expanding, you are done.
Do not simplify unnecessarily — stop the moment you match an option.
Why this question: Tests whether you recognize that x/|x| with a domain restriction collapses to a constant.
Solving path: For x < 0, replace |x| with -x. Then x/|x| = x/(-x) = -1. The derivative of any constant is 0. Do not attempt to differentiate x/|x| symbolically without this substitution first — you will get the wrong answer or waste time.
Why this question: The most direct chain rule test NDA uses — linear inner function makes it clean but the 3× multiplier is the trap.
Solving path: Outer function is sin, derivative is cos. Inner function is 3x + 2, derivative is 3. Chain rule: cos(3x + 2) × 3 = 3cos(3x + 2). Option A (cos(3x+2)) is the trap for forgetting the inner function's derivative.
Why this question: Classic quotient rule question where the sign in the numerator is the deciding factor.
Solving path: u = x+1, u' = 1; v = x-1, v' = 1. Numerator: 1·(x-1) - (x+1)·1 = x-1-x-1 = -2. Denominator: (x-1)². Answer: -2/(x-1)². If you write +2 in the numerator, you have swapped the subtraction — the QUOD pattern prevents this.
Why this question: Tests whether you apply the product rule correctly and whether you recognize equivalent factored forms.
Solving path: u = x³, u' = 3x²; v = eˣ, v' = eˣ. Product rule: 3x²eˣ + x³eˣ. Both options A (3x²eˣ + x³eˣ) and D (x²eˣ(3+x)) are the same expression. The question lists option A as correct — verify by factoring: x²eˣ(3+x) = 3x²eˣ + x³eˣ. Either form is acceptable, but confirm which option letter the question accepts.
Why this question: NDA tests non-differentiability — this is a concept question disguised as a calculation question.
Solving path: LHD at x=0: for x < 0, |x| = -x, so f'(x) = -1. RHD at x=0: for x > 0, |x| = x, so f'(x) = 1. LHD ≠ RHD, therefore f'(0) does not exist. This is not a limit question — it is a differentiability check. The most common wrong answer is 0 (students average -1 and 1). That is not how derivatives work.
Why this question: Tests recall of the inverse trig derivative formula — and whether you know which one has √(1-x²) versus 1+x².
Solving path: d/dx[sin⁻¹(x)] = 1/√(1-x²). This can be verified via implicit differentiation: let y = sin⁻¹x, then sin y = x. Differentiate: cos y · dy/dx = 1, so dy/dx = 1/cos y. Since cos y = √(1-sin²y) = √(1-x²), the answer is 1/√(1-x²). Option B (-1/√(1-x²)) is the cos⁻¹x derivative — a deliberate trap.
Why this question: Second product rule question, but with a trig function — tests whether you keep cos(x) and sin(x) in the right terms.
Solving path: u = x², u' = 2x; v = sin x, v' = cos x. Product rule: 2x·sin x + x²·cos x. Option C (x²cos x - 2xsin x) flips the sign and term order — this is the trap for students who mix up u'v - uv' (the quotient rule numerator) with u'v + uv' (the product rule).
Why this question: Direct recall question — but also derivable from the quotient rule, which NDA expects you to know.
Solving path: Write tan x = sin x / cos x. Quotient rule: numerator is cos x · cos x - sin x · (-sin x) = cos²x + sin²x = 1. Denominator is cos²x. So d/dx[tan x] = 1/cos²x = sec²x. Even if you have this memorized, knowing the derivation means you can reconstruct it mid-exam if you blank.
Forgetting the chain rule multiplier. Differentiating sin(3x+2) as just cos(3x+2) is wrong — you must multiply by 3, the derivative of the inner function. Every composition requires that extra multiplication, no exceptions.
Swapping the subtraction in the quotient rule. The numerator is u'v - uv', not uv' - u'v. Getting this wrong flips the sign of the entire answer. Write the rule down explicitly in the margin before applying it if you are not 100% certain.
Differentiating |x| at x = 0 as if it were smooth. The function |x| has a corner at the origin. Its derivative does not exist there. Averaging the left and right derivatives gives 0, but that is not a valid derivative — it is an arithmetic trick with no meaning.
Confusing sin⁻¹x and cos⁻¹x derivatives. The derivative of cos⁻¹x carries a minus sign: -1/√(1-x²). NDA answer choices will always include both options. Use the pair logic: their derivatives are negatives of each other because sin⁻¹x + cos⁻¹x = π/2 (a constant), so their derivatives must sum to zero.
Not checking whether the question specifies a domain. For piecewise or modulus functions, the domain restriction changes the function entirely. x/|x| for x < 0 is not the same problem as x/|x| in general. Read the domain condition every time before differentiating.
Applying the product rule when the simpler approach is expansion. For y = (x+1)(x-1) = x²-1, differentiate directly as 2x. Setting up the product rule here wastes 20 seconds. Expand first when the product is a simple binomial product.