Application of Derivatives

Application of Derivatives Class 12 NCERT Solutions Explained

By ConceptScroll Team · Published on 19 June 2026 · 5 min read

The application of derivatives class 12 NCERT solutions help students grasp how derivatives solve real-life problems in mathematics. This guide covers important concepts, formulas, and examples to aid Class 12 students in mastering this chapter effectively.

Understanding the Basics of Derivatives in Class 12

Derivatives represent the rate at which a function changes with respect to its variable. In Class 12 NCERT Mathematics, the derivative of a function $f(x)$ is denoted as $f'(x)$ or $\frac{dy}{dx}$. It helps in understanding how quantities vary and is fundamental for solving problems involving motion, growth, and optimisation.

Key concepts include:

  • Definition of derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

  • Interpretation: Slope of the tangent to the curve at any point
  • Basic derivative formulas:
  • $\frac{d}{dx} (x^n) = nx^{n-1}$
  • $\frac{d}{dx} (\sin x) = \cos x$

Mastering these basics is essential before moving to applications.

How Derivatives Are Applied to Find Increasing and Decreasing Intervals

One important application of derivatives is to determine where a function is increasing or decreasing. This helps in sketching graphs and understanding behaviour.

Steps to find intervals:

1. Find $f'(x)$. 2. Solve $f'(x) = 0$ to find critical points. 3. Test values around critical points to check the sign of $f'(x)$.

  • If $f'(x) > 0$, the function is increasing.
  • If $f'(x) < 0$, the function is decreasing.

Example:

Find intervals where $f(x) = x^3 - 3x^2 + 4$ is increasing or decreasing.

  • $f'(x) = 3x^2 - 6x = 3x(x - 2)$
  • Critical points: $x = 0, 2$
  • Test intervals:
  • For $x < 0$, say $x = -1$, $f'(-1) = 3(-1)(-3) = 9 > 0$ (increasing)
  • For $0 < x < 2$, say $x = 1$, $f'(1) = 3(1)(-1) = -3 < 0$ (decreasing)
  • For $x > 2$, say $x = 3$, $f'(3) = 3(3)(1) = 9 > 0$ (increasing)

Hence, $f(x)$ is increasing on $(-\infty, 0)$ and $(2, \infty)$, decreasing on $(0, 2)$.

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Using Derivatives to Identify Maxima and Minima

Derivatives help find local maxima and minima of functions, which are critical in optimisation problems.

Procedure:

1. Find $f'(x)$ and solve $f'(x) = 0$ to get critical points. 2. Use the second derivative test:

  • If $f''(x) > 0$ at a critical point, it is a local minimum.
  • If $f''(x) < 0$, it is a local maximum.
  • If $f''(x) = 0$, the test is inconclusive.

Example:

Find local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$.

  • $f'(x) = 3x^2 - 12x + 9$
  • Set $f'(x) = 0$: $3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0$
  • Roots: $x = 1, 3$
  • $f''(x) = 6x - 12$
  • At $x=1$, $f''(1) = 6 - 12 = -6 < 0$ (local maximum)
  • At $x=3$, $f''(3) = 18 - 12 = 6 > 0$ (local minimum)

This helps in graphing and solving real-world problems like cost minimisation.

Finding Tangents and Normals Using Derivatives

Derivatives give the slope of the tangent to a curve at any point, which is essential in geometry and physics.

  • Slope of tangent at $x=a$: $m = f'(a)$
  • Equation of tangent:

$$y - f(a) = f'(a)(x - a)$$

  • Slope of normal: $-\frac{1}{f'(a)}$
  • Equation of normal:

$$y - f(a) = -\frac{1}{f'(a)}(x - a)$$

Example:

Find the equation of the tangent and normal to $y = x^2$ at $x=2$.

  • $f'(x) = 2x$, so $f'(2) = 4$
  • Point: $(2, 4)$
  • Tangent: $y - 4 = 4(x - 2) \Rightarrow y = 4x - 4$
  • Normal slope: $-\frac{1}{4}$
  • Normal: $y - 4 = -\frac{1}{4}(x - 2) \Rightarrow y = -\frac{1}{4}x + \frac{9}{2}$

This application is vital for understanding curves and their properties.

Derivatives in Rate of Change Problems

Derivatives are used to calculate how one quantity changes with respect to another, commonly time.

Example:

If the radius $r$ of a sphere increases at $0.03$ cm/s, find the rate of change of volume when $r=10$ cm.

  • Volume of sphere: $V = \frac{4}{3} \pi r^3$
  • Differentiate w.r.t time $t$:

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$

  • Given $\frac{dr}{dt} = 0.03$ cm/s, $r=10$ cm
  • Substitute:

$$\frac{dV}{dt} = 4 \pi (10)^2 (0.03) = 4 \pi \times 100 \times 0.03 = 12 \pi \text{ cm}^3/s$$

This shows how derivatives help solve dynamic problems involving changing quantities.

Comparing First and Second Derivatives in Applications

Understanding the roles of first and second derivatives is crucial:

AspectFirst Derivative ($f'(x)$)Second Derivative ($f''(x)$)
PurposeRate of change, slope, increasing/decreasingConcavity, points of inflection, acceleration
Use in applicationsFind maxima/minima, monotonicityIdentify concavity, confirm nature of extrema
Graph interpretationSlope of tangentCurvature of graph

For example, $f'(x)$ helps locate critical points, while $f''(x)$ confirms if these points are maxima or minima.

Frequently asked questions

What is the importance of application of derivatives in Class 12 NCERT?

It helps solve real-world problems like finding maxima, minima, and rates of change.

How do I find local maxima and minima using derivatives?

Find where $f'(x)=0$ and use $f''(x)$ to check if points are maxima or minima.

Can derivatives help in sketching graphs of functions?

Yes, derivatives show increasing/decreasing intervals and concavity for graph sketching.

What is the difference between first and second derivatives in applications?

First derivative shows slope and rate of change; second derivative shows concavity and acceleration.

Are NCERT solutions enough to master the application of derivatives chapter?

Yes, practicing NCERT solutions builds strong conceptual understanding and problem-solving skills.

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