In Class 12 Mathematics, "Application of Derivatives" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
How do aeronautical engineers shape jet fuel tanks to maximize fuel capacity while using the minimum weight of aluminum, or how do quantitative financial algorithms predict the exact millisecond when a surging stock price reaches its peak? Optimization via calculus drives engineering.
Why This Chapter Matters
In Class 12 Mathematics, "Application of Derivatives" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
Derivatives and chain rule from Chapter 4.
Tangent lines.
Quadratic functions.
What You Will Learn (Core Objectives)
Calculate Rate of Change of quantities: $\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}$.
Determine intervals on which a function is Strictly Increasing ($f'(x) > 0$) or Strictly Decreasing ($f'(x) < 0$).
Apply First Derivative Test and Second Derivative Test to find Local Maxima, Local Minima, and Points of Inflection.
Find Absolute Maximum and Absolute Minimum values of a continuous function on a closed interval $[a, b]$.
Solve real-world optimization word problems in mensuration and business.
Chapter Roadmap & Progression
11. Rate of Change & Monotonicity
22. Maxima & Minima (Optimization Te...
33. Absolute Extrema on Closed Inter...
Complete Concept Guide (100% Curriculum Coverage)
1. Rate of Change & Monotonicity
Rate of Change: $\frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}$. (E.g. Expanding circular water ripples).
A critical point occurs where $\mathbf{f'(c) = 0}$ or $f'(c)$ does not exist. • Second Derivative Test: 1. If $\mathbf{f''(c) < 0}$, then $x = c$ is a point of Local Maximum. 2. If $\mathbf{f''(c) > 0}$, then $x = c$ is a point of Local Minimum. 3. If $f''(c) = 0$, test fails → use First Derivative sign change test!
3. Absolute Extrema on Closed Intervals
To find absolute maximum/minimum on $[a, b]$: evaluate function values at all critical points in $(a, b)$ AND at endpoints $a$ and $b$. The highest value is the Absolute Maximum; lowest is Absolute Minimum.
Application of Derivatives - Key Conceptual & Analytical Model
Critical Points: Roots of $f'(x) = 0$ identifying candidate peaks and valleys.
Takeaway 4
Second Derivative Curvature: Negative second derivative indicates concave-down peak.
Takeaway 5
Closed Interval Extrema: Comparing critical point values against boundary endpoints.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
The radius of a circle is increasing uniformly at the rate of $3\text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $10\text{ cm}$.
Reveal Answer & Explanation
Answer: Area $A = \pi r^2$. $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. Given $\frac{dr}{dt} = 3\text{ cm/s}$ and $r = 10\text{ cm}$: $\frac{dA}{dt} = 2\pi(10)(3) = 60\pi\text{ cm}^2\text{/s} \approx 188.5\text{ cm}^2\text{/s}$. 60π cm²/s.
2
Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing.
Reveal Answer & Explanation
Answer: $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$. Critical points: $x = -2, 3$. Intervals: $(-\infty, -2)$ has $f'(x) > 0$ (Strictly Increasing); $(-2, 3)$ has $f'(x) < 0$ (Strictly Decreasing); $(3, \infty)$ has $f'(x) > 0$ (Strictly Increasing). Increasing on (-∞, -2) U (3, ∞); Decreasing on (-2, 3).
3
Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.
Reveal Answer & Explanation
Answer: Let numbers be $x$ and $15 - x$. $S(x) = x^2 + (15 - x)^2 = 2x^2 - 30x + 225$. $S'(x) = 4x - 30 = 0 \implies x = 7.5$. $S''(x) = 4 > 0$ (Minimum). The two numbers are both 7.5. Numbers are 7.5 and 7.5.
4
Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $8/27$ of the volume of the sphere.
Reveal Answer & Explanation
Answer: Let cone height be $h$ and base radius $r$. From sphere geometry: $r^2 = R^2 - (h - R)^2 = 2Rh - h^2$. Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2Rh^2 - h^3)$. $\frac{dV}{dh} = \frac{\pi}{3}(4Rh - 3h^2) = 0 \implies h = \frac{4}{3}R$. $V_{\text{max}} = \frac{1}{3}\pi [2R(\frac{16R^2}{9}) - \frac{64R^3}{27}] = \frac{32}{81}\pi R^3 = \frac{8}{27}\left(\frac{4}{3}\pi R^3\right) = \frac{8}{27} V_{\text{sphere}}$. Proved V_cone = 8/27 V_sphere.
5
Find the absolute maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.
Reveal Answer & Explanation
Answer: $f'(x) = 6x^2 - 30x + 36 = 6(x - 2)(x - 3) = 0 \implies x = 2, 3$. Evaluate: $f(1) = 24$, $f(2) = 29$, $f(3) = 28$, $f(5) = 56$. Absolute Maximum is 56 (at $x=5$); Absolute Minimum is 24 (at $x=1$). Abs Max = 56 (at x=5), Abs Min = 24 (at x=1).
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