Free, step-by-step NCERT Solutions for all four parts of this chapter — rate of change of quantities, increasing and decreasing functions, maxima and minima, and absolute maximum/minimum on a closed interval. Solved the way CBSE awards marks, with the key formulas and the mistakes that cost students marks every year, right on this page.
This Class 12 Maths NCERT Solutions Chapter 6 hub covers Application of Derivatives, which takes the differentiation techniques from Chapter 5 and turns them into a practical toolkit. You'll use the derivative to measure how fast one quantity changes as another one does, to work out where a function is rising or falling, and — most importantly for the board exam — to find the largest or smallest value a function can take, whether that's the maximum area you can enclose, the minimum material needed for a box, or the shortest possible distance between two points.
The chapter builds in a deliberate sequence: rate of change first (pure differentiation, applied), then increasing/decreasing behaviour (reading the sign of the derivative), then maxima and minima (using that sign behaviour to locate turning points), and finally absolute maxima and minima on a closed interval, which is what most real optimisation problems — the ones that show up as 5-mark or case-study questions — actually need.
dy/dx tells you how fast y changes as x changes. Exercise 6.1.
The sign of f′(x) tells you whether the function is rising or falling. Exercise 6.2.
Where f′(x) changes sign (or f″(c) tells you the curve's shape), you get a turning point. Exercise 6.3.
On a closed interval, compare critical points with the endpoints to find the true largest/smallest value. Misc. Exercise.
Everything you need before you start solving. This is a summary for quick recall — the Formula Deck below has the full printable version for all of Calculus.
Rate of change of y at a specific point: \left.\dfrac{dy}{dx}\right|_{x=x_0}
Essential whenever both x and y depend on time t.
Nearly every question in Ex 6.1 hides a geometry formula — keep these on hand.
Strictly increasing if f′(x) > 0 throughout I.
Strictly decreasing if f′(x) < 0 throughout I.
Necessary condition for a local max/min — not sufficient by itself.
Works even where f is not twice differentiable.
If f″(c) = 0 too, the test fails — go back to the first derivative test.
Missing the endpoints is the single most common way marks are lost here.
A quick way to decide, once you've found where f′(x) = 0.
| Situation | Use this | Why |
|---|---|---|
| f is easy to differentiate twice, and f″(c) ≠ 0 | Second Derivative Test | Fastest — one substitution tells you max or min. |
| f″(c) = 0, or f isn't differentiable at c | First Derivative Test | Second derivative test explicitly fails here — sign-check f′(x) on either side instead. |
| The problem gives a closed interval [a, b] | Absolute Max/Min working rule | Local tests alone can miss the true maximum, which may sit at an endpoint. |
| You just need to prove f is increasing/decreasing (no turning point needed) | Sign of f′(x) directly | No need to locate critical points at all for a pure increasing/decreasing proof. |
Drawn from where students actually lose marks across all four exercises.
Rate of change of quantities — circles, cubes, spheres, cones, marginal cost & revenue · 18 questions
Solve Exercise 6.1 →Increasing and decreasing functions — proving monotonicity and finding intervals · 19 questions
Solve Exercise 6.2 →Maxima and minima — local extrema, first & second derivative test, absolute max/min, optimisation · 29 questions
Solve Exercise 6.3 →Mixed application questions combining rate of change, monotonicity and optimisation · 16 questions
Solve Miscellaneous →Every formula for Calculus — differentiation, applications of derivatives, integration — in one printable PDF.
Get Formula Deck →The AI Question Bank has board-tagged MCQs, Assertion-Reason and Case Studies for Application of Derivatives, with instant feedback.
Explore AI Q-Bank →Quick answers about Chapter 6, Application of Derivatives.
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