Why applications outweigh the rules
Differentiating a function is a mechanical skill that most students acquire quickly. Interpreting the result is not. Knowing that f′ > 0 means increasing, that f′ = 0 marks a stationary point, and that f″ distinguishes a maximum from a minimum turns the derivative from an answer into a tool.
That is where the exam concentrates. A question asking for the interval on which a function increases, or the maximum volume of a box cut from a fixed sheet, tests the interpretation - the differentiation inside it is a single line.
How JEE actually asks Differentiation
NTA does not publish chapter-wise weightage, so treat any figure online as a coaching estimate from past papers, and confirm your session's syllabus and pattern in the official NTA information bulletin.
The observable pattern is that applications of derivatives supply more questions than raw differentiation. Tangents and normals, monotonicity, maxima and minima, and rates of change recur constantly, often as word problems requiring you to build the function first. Rolle's theorem and the mean value theorem appear as short conceptual items.
Key concepts, compressed
- ✓The derivative is the instantaneous rate of change, and geometrically the slope of the tangent.
- ✓The chain rule handles composition: differentiate the outer function, then multiply by the derivative of the inner one.
- ✓Implicit differentiation treats y as a function of x, so differentiating y² gives 2y(dy/dx).
- ✓Logarithmic differentiation is for variable exponents and long products - take logs first, then differentiate.
- ✓f′ > 0 means increasing, f′ < 0 decreasing. Stationary points have f′ = 0.
- ✓f″ > 0 at a stationary point means a local minimum, f″ < 0 a local maximum. If f″ = 0 the test is inconclusive.
Rules you need before attempting the questions
| Rule | Formula | Note |
|---|---|---|
| Power rule | d/dx xⁿ = n x^(n−1) | |
| Product rule | (uv)′ = u′v + uv′ | |
| Quotient rule | (u/v)′ = (u′v − uv′)/v² | order matters |
| Chain rule | dy/dx = (dy/du)(du/dx) | |
| Parametric | dy/dx = (dy/dt)/(dx/dt) | |
| d/dx sin x | cos x | |
| d/dx e^x | e^x | |
| d/dx ln x | 1/x | |
| d/dx tan⁻¹x | 1/(1 + x²) | |
| Slope of tangent | dy/dx at the point | |
| Slope of normal | −1/(dy/dx) | |
| Second derivative test | f″ > 0 min, f″ < 0 max | inconclusive at 0 |
The five mistakes that cost the most marks
- ✓Forgetting the inner derivative in the chain rule. This is the most frequent single error in the entire chapter.
- ✓Reversing the quotient rule numerator. It is u′v − uv′; writing uv′ − u′v negates every answer.
- ✓Testing only f′ = 0 for extrema. Critical points also occur where the derivative is undefined, and on a closed interval the endpoints must be evaluated too.
- ✓Concluding an inflection point from f″ = 0. The function x⁴ has f″(0) = 0 at a genuine minimum - the second derivative must change sign.
- ✓Confusing tangent and normal slopes. They are negative reciprocals, and questions deliberately give one and ask for the other.
Practice set 1: the rules
1. Differentiate y = x⁵.
dy/dx = 5x⁴. The power rule brings the exponent down as a coefficient and reduces it by one. It holds for all real exponents, not just positive integers - so x^(1/2) differentiates to (1/2)x^(−1/2), and x^(−1) to −x^(−2).
2. Differentiate y = sin x.
dy/dx = cos x, with x in radians. In degrees an extra factor of π/180 appears, which is why calculus is always done in radians. The companion results are that cos x differentiates to −sin x, with the minus sign, and tan x to sec²x.
3. Differentiate y = x² sin x.
dy/dx = 2x sin x + x² cos x. This is a product, so apply u′v + uv′ with u = x² and v = sin x. A useful check on any product-rule answer is that it must contain two terms - a single term means one part of the rule was dropped.
4. Differentiate y = x/(x + 1).
dy/dx = 1/(x + 1)². By the quotient rule, the numerator is (1)(x + 1) − (x)(1) = x + 1 − x = 1, over (x + 1)². Note that the result is positive everywhere it is defined, so the function is increasing on both sides of its vertical asymptote at x = −1.
5. Differentiate y = sin 3x.
dy/dx = 3 cos 3x. The chain rule requires multiplying by the derivative of the inner function 3x, which is 3. Answering cos 3x is the single most common mistake in the chapter, and it will always appear among the options.
6. Differentiate y = e^(2x).
dy/dx = 2e^(2x). Again the chain rule: the derivative of the inner function 2x is 2. The general result d/dx e^(kx) = k e^(kx) is worth having directly, since exponentials with linear exponents appear constantly in differential equations later.
7. Differentiate y = ln x, and state the domain of the result.
dy/dx = 1/x, valid for x > 0 since ln x is only defined there. If the question uses ln|x| instead, the derivative is still 1/x but now valid for all x ≠ 0 - which is exactly why the integral of 1/x is written with an absolute value.
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Practice set 2: implicit, parametric and logarithmic
8. Find dy/dx for x² + y² = 25.
dy/dx = −x/y. Differentiate both sides with respect to x, treating y as a function of x: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. Implicit differentiation is required whenever the relation cannot conveniently be solved for y, and the giveaway is the dy/dx factor appearing whenever a y is differentiated.
9. For x = t² and y = t³, find dy/dx.
dy/dx = 3t/2. Compute dy/dt = 3t² and dx/dt = 2t, then divide: 3t²/2t = 3t/2 for t ≠ 0. Note the answer is left in terms of the parameter, which is normal and expected. Attempting to eliminate t first is usually slower and sometimes impossible.
10. Differentiate y = x^x.
dy/dx = x^x(1 + ln x). The variable appears in both the base and the exponent, so neither the power rule nor the exponential rule applies. Take logarithms: ln y = x ln x. Differentiating gives (1/y)(dy/dx) = ln x + 1, so dy/dx = y(1 + ln x) = x^x(1 + ln x). Any time the exponent contains x, logarithmic differentiation is the route.
11. What is the second derivative of y = x³ − 3x?
d²y/dx² = 6x. Differentiate once to get 3x² − 3, then again to get 6x. The second derivative measures how the slope is changing, which is what makes it the tool for distinguishing maxima from minima - and this particular function is used in questions 19 to 21.
12. Differentiate y = tan⁻¹x.
dy/dx = 1/(1 + x²). This is a standard inverse-trigonometric derivative and it appears constantly in integration, where the reverse direction turns 1/(1 + x²) into an arctangent. The related result is that sin⁻¹x differentiates to 1/√(1 − x²).
13. When should you use logarithmic differentiation?
When the variable appears in an exponent, or when the function is a long product or quotient. Taking logarithms converts products to sums, quotients to differences and exponents to coefficients, which usually turns several applications of the product rule into one short calculation.
Practice set 3: tangents and normals
14. Find the equation of the tangent to y = x² at the point (2, 4).
y = 4x − 4. The slope is dy/dx = 2x, which at x = 2 gives 4. Using the point-slope form, y − 4 = 4(x − 2), which simplifies to y = 4x − 4. Always evaluate the derivative at the point rather than leaving it as a function - the tangent needs a number, not an expression.
15. Find the slope of the normal to that curve at the same point.
−1/4. The normal is perpendicular to the tangent, so its slope is the negative reciprocal of 4. Its equation would be y − 4 = −(1/4)(x − 2). A question that gives you the normal's slope and asks for the curve works backwards through the same relation.
16. Where does the curve y = x³ − 3x have horizontal tangents?
At x = 1 and x = −1. A horizontal tangent means zero slope, so set dy/dx = 3x² − 3 = 0, giving x² = 1. The corresponding points are (1, −2) and (−1, 2). These are the stationary points, and the next set determines which is a maximum and which a minimum.
17. How do you find the angle between two curves at their point of intersection?
Find the point of intersection, compute each curve's slope there, and take the angle between the two lines using tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. The curves are orthogonal when m₁m₂ = −1, which is a condition questions often ask you to verify rather than compute.
18. What is special about the point (0, 0) on the curve y = x³?
It is a stationary point that is neither a maximum nor a minimum - it is a point of inflection with a horizontal tangent. Here f′(0) = 0 and f″(0) = 0, so the second derivative test fails, and you must examine the sign of f′ on either side. It is positive on both, so the function is increasing throughout and simply flattens momentarily.
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Practice set 4: monotonicity, maxima and minima
19. Find the critical points of f(x) = x³ − 3x.
x = 1 and x = −1. Set f′(x) = 3x² − 3 = 0, so x² = 1. Since f′ is a polynomial it is defined everywhere, so there are no additional critical points from undefined derivatives - but on a function with a fraction or a root, those must be checked too.
20. Classify those critical points using the second derivative test.
x = 1 is a local minimum and x = −1 is a local maximum. With f″(x) = 6x: at x = 1, f″ = 6 which is positive, so the curve is concave up and the point is a minimum. At x = −1, f″ = −6, negative and concave down, giving a maximum. The mnemonic worth keeping is that a positive second derivative holds water.
21. What are the local maximum and minimum values of that function?
A local maximum of 2 at x = −1 and a local minimum of −2 at x = 1. Substituting: f(−1) = −1 + 3 = 2 and f(1) = 1 − 3 = −2. Note the counter-intuitive feature - the local maximum value here is greater than the local minimum, but neither is a global extremum, since the cubic tends to plus and minus infinity.
22. On what interval is f(x) = x³ − 3x increasing?
For x < −1 and x > 1. The function increases where f′ > 0, that is 3x² − 3 > 0, which gives x² > 1. Between −1 and 1 the derivative is negative and the function decreases. Solving the inequality rather than testing sample points is the reliable method, especially when the derivative has several roots.
23. Besides f′ = 0, where else can a critical point occur?
Wherever f′ is undefined but f itself is defined. The function |x| has a critical point at x = 0 for exactly this reason, and x^(2/3) has one there too. Restricting your search to f′ = 0 misses these entirely, which is why a question involving a modulus or a fractional power is testing this specific gap.
24. How do you find the absolute maximum of a continuous function on a closed interval?
Evaluate the function at every critical point inside the interval and at both endpoints, then compare the values. The second derivative test only classifies local behaviour and says nothing about endpoints - and for many functions the absolute extremum sits at an endpoint where the derivative is not zero at all.
25. Does f″(x) = 0 guarantee a point of inflection?
No. The second derivative must actually change sign there. For f(x) = x⁴, f″(0) = 0 but f″ = 12x² is non-negative everywhere, so the concavity never changes and x = 0 is a minimum rather than an inflection. Checking the sign on both sides is the necessary extra step.
Practice set 5: rates of change and theorems
26. The radius of a sphere increases at 2 cm/s. How fast is the volume increasing when the radius is 5 cm?
200π cm³/s. Differentiate V = (4/3)πr³ with respect to time: dV/dt = 4πr²(dr/dt). Substituting r = 5 and dr/dt = 2 gives 4π(25)(2) = 200π cm³/s. The structure is always the same - differentiate the geometric relation with respect to t, then substitute the instantaneous values last, not first.
27. What is the general method for a related-rates problem?
Write the relation between the quantities, differentiate the whole relation with respect to time, then substitute the given instantaneous values. The most common error is substituting the numerical values before differentiating, which freezes a varying quantity into a constant and makes its rate vanish.
28. State the conditions for Rolle's theorem.
The function must be continuous on the closed interval [a, b], differentiable on the open interval (a, b), and satisfy f(a) = f(b). Then there is at least one c in (a, b) with f′(c) = 0. Questions usually supply a function and ask whether the theorem applies - and the answer is often no, because differentiability fails somewhere inside.
29. How does the mean value theorem differ from Rolle's theorem?
It drops the requirement that f(a) = f(b). The conclusion becomes f′(c) = (f(b) − f(a))/(b − a) - there is a point where the instantaneous rate equals the average rate over the interval. Rolle's theorem is the special case where that average rate is zero.
30. Use differentials to approximate the change in y = x² when x changes from 3 to 3.01.
About 0.06. The differential dy = 2x dx, so with x = 3 and dx = 0.01 we get dy = 6 × 0.01 = 0.06. The exact change is 3.01² − 3² = 9.0601 − 9 = 0.0601, so the approximation is accurate to four decimal places. This linear approximation is exactly what the tangent line provides, and its error grows with the square of dx.
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How to study this chapter efficiently
- ✓Say the chain rule aloud as you apply it - outer derivative times inner derivative. The forgotten inner factor is the chapter's most expensive habit.
- ✓Check every product-rule answer has two terms and every quotient-rule numerator starts with u′v. Both checks take a second.
- ✓Weight your practice toward applications, not rules. That is where the paper puts its questions.
- ✓For every extremum problem, list critical points from f′ = 0, from f′ undefined, and from the endpoints - as three separate lines.
- ✓When the second derivative test gives zero, fall back to the sign of f′ on either side rather than guessing.
- ✓For word problems, write the relation before differentiating anything, and substitute numbers only at the very end.
Turn this into active practice
Reading a worked maxima-and-minima solution is deceptive: the function is already built, so the hardest step has been done for you. The exam gives you a sheet of metal and a fold, and expects the function to come from you.
The JEE Differentiation quiz on QUFF generates fresh questions across the rules, implicit and parametric forms, tangents, monotonicity and optimisation, marks them instantly and explains each answer. Do a mixed timed set, and separate your errors into rule slips and interpretation errors - the first needs drilling, the second needs more varied problems.
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