Anand Classes provides the most detailed and accurate Limits and Derivatives Exercise 12.2 NCERT Solutions for Class 11 (Set-1) of Chapter – 12, designed to help students master important calculus concepts with step-by-step explanations. These solutions are based on the latest CBSE syllabus and NCERT textbook, making them ideal for exam preparation and concept clarity. Whether you’re revising for school tests or board exams, these Class 11 Maths solutions make learning easier and more effective. Click the print button to download study material and notes.
NCERT Question.1 : Find the derivative of $x^2 – 2$ at $x = 10$
Solution:
Let
$$f(x) = x^2 – 2$$
Then,
$$f(x + h) = (x + h)^2 – 2$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
When $x = 10$,
$$f'(10) = \lim_{h \to 0} \frac{f(10 + h) – f(10)}{h}$$
Substitute the values:
$$f'(10) = \lim_{h \to 0} \frac{[(10 + h)^2 – 2] – [10^2 – 2]}{h}$$
Simplify the numerator:
$$f'(10) = \lim_{h \to 0} \frac{(100 + 20h + h^2 – 2) – (100 – 2)}{h}$$
$$f'(10) = \lim_{h \to 0} \frac{(20h + h^2)}{h}$$
$$f'(10) = \lim_{h \to 0} (20 + h)$$
Now, as $h \to 0$,
$$f'(10) = 20$$
$$
\boxed{f'(10) = 20}
$$
This solution covers derivatives using first principles, ideal for CBSE Class 12 Mathematics, JEE preparation, and calculus concept building.
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NCERT Question.2 : Find the derivative of $x$ at $x = 1$
Solution:
Let
$$f(x) = x$$
Then,
$$f(x + h) = x + h$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
When $x = 1$,
$$f'(1) = \lim_{h \to 0} \frac{f(1 + h) – f(1)}{h}$$
Substitute the values:
$$f'(1) = \lim_{h \to 0} \frac{(1 + h) – 1}{h}$$
Simplify:
$$f'(1) = \lim_{h \to 0} \frac{h}{h}$$
$$f'(1) = \lim_{h \to 0} 1$$
Hence,
$$f'(1) = 1$$
$$
\boxed{f'(1) = 1}
$$
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NCERT Question.3 : Find the derivative of $99x$ at $x = 100$
Solution:
Let
$$f(x) = 99x$$
Then,
$$f(x + h) = 99(x + h)$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
When $x = 100$,
$$f'(100) = \lim_{h \to 0} \frac{f(100 + h) – f(100)}{h}$$
Substitute the values:
$$f'(100) = \lim_{h \to 0} \frac{[99(100 + h)] – [99(100)]}{h}$$
Simplify the numerator:
$$f'(100) = \lim_{h \to 0} \frac{(9900 + 99h – 9900)}{h}$$
$$f'(100) = \lim_{h \to 0} \frac{99h}{h}$$
$$f'(100) = \lim_{h \to 0} 99$$
Hence,
$$f'(100) = 99$$
$$
\boxed{f'(100) = 99}
$$
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NCERT Question 4.1 : Find the derivative of $x^3 – 27$ from first principle.
Solution:
Let
$$f(x) = x^3 – 27$$
Then,
$$f(x + h) = (x + h)^3 – 27$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
Substitute the values:
$$f'(x) = \lim_{h \to 0} \frac{[(x + h)^3 – 27] – [x^3 – 27]}{h}$$
Simplify:
$$f'(x) = \lim_{h \to 0} \frac{(x + h)^3 – x^3}{h}$$
Expand $(x + h)^3$:
$$f'(x) = \lim_{h \to 0} \frac{[x^3 + 3x^2h + 3xh^2 + h^3 – x^3]}{h}$$
Simplify the numerator:
$$f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$
Divide each term by $h$:
$$f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)$$
Now, as $h \to 0$,
$$f'(x) = 3x^2$$
$$
\boxed{f'(x) = 3x^2}
$$
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NCERT Question 4.2 : Find the derivative of $(x – 1)(x – 2)$ from first principle
Solution:
Let
$$f(x) = (x – 1)(x – 2)$$
Then,
$$f(x + h) = (x + h – 1)(x + h – 2)$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
Substitute the values:
$$f'(x) = \lim_{h \to 0} \frac{[(x + h – 1)(x + h – 2)] – [(x – 1)(x – 2)]}{h}$$
Expand $(x + h – 1)(x + h – 2)$:
$$(x + h – 1)(x + h – 2) = x^2 + 2hx + h^2 – 3x – 3h + 2$$
So,
$$f'(x) = \lim_{h \to 0} \frac{[x^2 + 2hx + h^2 – 3x – 3h + 2] – [x^2 – 3x + 2]}{h}$$
Simplify the numerator:
$$f'(x) = \lim_{h \to 0} \frac{(2hx + h^2 – 3h)}{h}$$
Divide each term by $h$:
$$f'(x) = \lim_{h \to 0} (2x + h – 3)$$
Now, as $h \to 0$,
$$f'(x) = 2x – 3$$
$$
\boxed{f'(x) = 2x – 3}
$$
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NCERT Question.4.3 : Find the derivative of $\dfrac{1}{x^2}$ from first principle
Solution:
Let
$$f(x) = \dfrac{1}{x^2}$$
Then,
$$f(x + h) = \dfrac{1}{(x + h)^2}$$
From the first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
Substitute the values:
$$f'(x) = \lim_{h \to 0} \dfrac{\dfrac{1}{(x + h)^2} – \dfrac{1}{x^2}}{h}$$
Simplify the numerator using the LCM $x^2 (x + h)^2$:
$$f'(x) = \lim_{h \to 0} \frac{x^2 – (x + h)^2}{h x^2 (x + h)^2}$$
Expand $(x + h)^2$:
$$f'(x) = \lim_{h \to 0} \frac{x^2 – [x^2 + 2xh + h^2]}{h x^2 (x + h)^2}$$
Simplify the numerator:
$$f'(x) = \lim_{h \to 0} \dfrac{-2xh – h^2}{h x^2 (x + h)^2}$$
Factor $h$ from the numerator:
$$f'(x) = \lim_{h \to 0} \dfrac{h(-2x – h)}{h x^2 (x + h)^2}$$
Cancel $h$:
$$f'(x) = \lim_{h \to 0} \frac{-2x – h}{x^2 (x + h)^2}$$
Now, as $h \to 0$:
$$f'(x) = \frac{-2x}{x^2 \cdot x^2}$$
Simplify:
$$f'(x) = -\frac{2}{x^3}$$
$$
\boxed{f'(x) = -\frac{2}{x^3}}
$$
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NCERT Question.4.4 : Find the derivative of $\dfrac{x+1}{x-1}$ from first principle
Solution:
Let
$$f(x)=\frac{x+1}{x-1}.$$
Then
$$f(x+h)=\frac{x+1+h}{x-1+h}.$$
From the first principle,
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
Compute the difference:
$$\frac{f(x+h)-f(x)}{h}
=\frac{1}{h}\left(\frac{x+1+h}{x-1+h}-\frac{x+1}{x-1}\right).$$
Combine the fractions in the numerator:
$$\frac{x+1+h}{x-1+h}-\frac{x+1}{x-1}
=\frac{(x+1+h)(x-1)-(x+1)(x-1+h)}{(x-1+h)(x-1)}.$$
Expand and simplify the numerator:
$$(x+1+h)(x-1)-(x+1)(x-1+h)$$
$$ =[(x+1)(x-1)+h(x-1)]-[(x+1)(x-1)+h(x+1)]$$
$$=h(x-1)-h(x+1)=h\big((x-1)-(x+1)\big)=h(-2)=-2h.$$
Thus
$$\frac{f(x+h)-f(x)}{h}=\frac{-2h}{h(x-1+h)(x-1)}=\frac{-2}{(x-1+h)(x-1)}.$$
Taking the limit $h\to 0$:
$$f'(x)=\lim_{h\to 0}\frac{-2}{(x-1+h)(x-1)}=\frac{-2}{(x-1)^2}.$$
$$
\boxed{f'(x)=-\dfrac{2}{(x-1)^2}}
$$
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NCERT Question.5 : For the function
$$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^{2}}{2}+x+1$$
prove that $f'(1) = 100 f'(0)$
Solution:
From termwise differentiation (power rule) for $n\ge 1$
$$\frac{d}{dx}\left(\frac{x^n}{n}\right)=x^{,n-1}$$
and $\dfrac{d}{dx}(x)=1,\ \dfrac{d}{dx}(1)=0$.
Hence
$$f'(x)=x^{99}+x^{98}+\cdots+x+1.$$
Evaluate at $x=1$:
$$f'(1)=1^{99}+1^{98}+\cdots+1+1=100.$$
Evaluate at $x=0$:
$$f'(0)=0^{99}+0^{98}+\cdots+0+1=1.$$
Therefore
$$f'(1)=100=100\cdot 1=100f'(0)$$
$$
\boxed{f'(1)=100 f'(0)}
$$
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Question 6. Find the derivative of $ x^n + a x^{n-1} + a^2 x^{n-2} + \ldots + a^{n-1}x + a^n $ for a fixed real number $a$.
Solution:
Given,
$$f(x) = x^n + a x^{n-1} + a^2 x^{n-2} + \ldots + a^{n-1}x + a^n$$
As the derivative of ( x^m ) is ( m x^{m-1} ), and the derivative of a constant is ( 0 ),
we differentiate term by term.
So,
$$
f'(x) = \frac{d}{dx}(x^n) + \frac{d}{dx}(a x^{n-1}) + \frac{d}{dx}(a^2 x^{n-2}) + \ldots + \frac{d}{dx}(a^{n-1}x) + \frac{d}{dx}(a^n) $$
$$ = n x^{n-1} + a (n-1) x^{n-2} + a^2 (n-2) x^{n-3} + \ldots + a^{n-2} (2x) + a^{n-1}(1) + 0 $$
Hence,
$$f'(x) = n x^{n-1} + a (n-1) x^{n-2} + a^2 (n-2) x^{n-3} + \ldots + 2a^{n-2}x + a^{n-1}$$
or, in summation form,
$$f'(x) = \sum_{k=0}^{n-1} a^k (n-k) x^{n-k-1}$$
$$
\boxed{f'(x) = n x^{n-1} + a (n-1) x^{n-2} + a^2 (n-2) x^{n-3} + \ldots + 2a^{n-2}x + a^{n-1}}
$$
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