Read Part I Chapter 03 Differentiation of MSBSHSE Class 12 Maths Commerce
Review the Maharashtra Board Class 12 Maths Commerce Part I Chapter 3 Differentiation PDF Download designed for Class 12 Maths Commerce students. Published under the latest MSBSHSE guidelines for 2026-27, this chapter-wise resource supports daily study and targeted revision.
Part I Chapter 03 Differentiation PDF Resource
Access the complete PDF for Part I Chapter 03 Differentiation below. This focused excerpt allows students to isolate specific topics for thorough review. Cross-reference your textbook exercises with our detailed MSBSHSE Solutions for Class 12 Maths Commerce.
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.
| \(y = f(x)\) | \(\frac{dy}{dx} = f'(x)\) | |
|---|---|---|
| 1 | K (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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Free study material for Maths Commerce
Official MSBSHSE Textbook PDF: Class 12 Maths Commerce Part I Chapter 03 Differentiation
Class 12 Maths Commerce Part I Chapter 03 Differentiation Official E-Book
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