Maharashtra Board Class 12 Maths Commerce Part I Chapter 3 Differentiation PDF Download

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MSBSHSE Class 12 Maths Commerce Part I Chapter 3 Differentiation Digital Edition

For Class 12 Maths Commerce, this chapter in Maharashtra Board Class 12 Maths Commerce Part I Chapter 3 Differentiation PDF Download provides a detailed overview of important concepts. We highly recommend using this text alongside the MSBSHSE Solutions for Class 12 Maths Commerce to learn the exercise questions provided at the end of the chapter.

Part I Chapter 3 Differentiation MSBSHSE Book Class 12 PDF (2026-27)

Differentiation

Let's Study

1. Derivatives of composite functions.

2. Derivatives of inverse functions.

3. Derivatives of logarithmic functions.

4. Derivatives of implicit function.

5. Derivatives of parametric functions.

6. Derivative of second order.

Let's Recall

1. Concept of continuity

2. Concept of Differentiability.

3. Derivatives of some standard functions.

\(y = f(x)\)\(\frac{dy}{dx} = f'(x)\)
1K (constant)0
2\(x\)1
3\(\sqrt{x}\)\(\frac{1}{2\sqrt{x}}\)
4\(\frac{1}{x}\)\(-\frac{1}{x^2}\)
5\(x^n\)\(n.x^{n-1}\)
6\(a^x\)\(a^x.\log a\)
7\(e^x\)\(e^x\)
8\(\log x\)\(\frac{1}{x}\)

Rules of Differentiation:

If u and v are differentiable functions of x and if

1. \(y = u + v\) then \(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\)

2. \(y = u - v\) then \(\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}\)

3. \(y = u.v\) then \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)

4. \(y = \frac{u}{v}\) then \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\), \(v \ne 0\)

5. \(y = k.u\) then \(\frac{dy}{dx} = k.\frac{du}{dx}\), k constant.

Introduction:

In Standard XI, we have studied the concept of differentiation. We have used this concept in calculating marginal demand and marginal cost of a commodity.

Let's Learn

3.1 Derivative of A Composite Function:

Sometimes complex looking functions can be greatly simplified by expressing them as compositions of two or more different functions. It is then not possible to differentiate them directly is possible with simple functions.

Now, we discuss differentiation of such composite functions using the chain rule.

Result 1: If \(y = f(u)\) is a differentiable function of u and \(u = g(x)\) is a differentiable function of x then

\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]

(This is called Chain Rule)

Teacher's Note

Chain rule helps us find derivatives of difficult functions. For example, if you need to find how fast a balloon grows when you pump air into it.

Exam Trick

Remember: Chain rule means find the derivative step by step. Like making tea: boil water, then add tea, then add milk. Do each step one after another.

Points to Remember

Chain rule works for functions inside functions.


First differentiate the outside function, then the inside function.


Always multiply the derivatives together.


This rule makes hard problems become easy.

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MSBSHSE Book Class 12 Maths Commerce Part I Chapter 3 Differentiation

Download the official MSBSHSE Textbook for Class 12 Maths Commerce Part I Chapter 3 Differentiation, updated for the latest academic session. These e-books are the main textbook used by major education boards across India. All teachers and subject experts recommend the Part I Chapter 3 Differentiation NCERT e-textbook because exam papers for Class 12 are strictly based on the syllabus specified in these books. You can download the complete chapter in PDF format from here.

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