Anand Classes presents comprehensive NCERT Solutions for Limits and Derivatives Exercise 12.2 Class 11 (Set-2) of Chapter 12 in PDF format, carefully prepared to help students understand fundamental calculus concepts with clarity. These step-by-step solutions follow the latest CBSE guidelines and NCERT textbook, making them perfect for Class 11 students aiming to strengthen their mathematical foundation and score high in exams. Each question is solved with detailed explanations to enhance conceptual understanding. Click the print button to download study material and notes.
NCERT Question 7.1 : For some constants $a$ and $b$, find the derivative of $(x – a)(x – b)$
Solution:
Let
$$f(x) = (x – a)(x – b)$$
Expanding, we get
$$f(x) = x^2 – (a + b)x + ab$$
Now, differentiating both sides with respect to $x$,
$$\frac{d}{dx} \big(f(x)\big) = \frac{d}{dx} \big(x^2 – (a + b)x + ab\big)$$
So,
$$f'(x) = \frac{d}{dx}(x^2) – \frac{d}{dx}((a + b)x) + \frac{d}{dx}(ab)$$
We know that:
- The derivative of $x^n$ is $n x^{n-1}$
- The derivative of a constant is $0$
Hence,
$$f'(x) = 2x – (a + b)(1) + 0$$
Simplifying, we get
$$f'(x) = 2x – (a + b)$$
$$
\boxed{f'(x) = 2x – a – b}
$$
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NCERT Question 7.2 : For some constants $a$ and $b$, find the derivative of $(ax^2 + b)^2$
Solution:
Let
$$f(x) = (ax^2 + b)^2$$
Expanding the expression,
$$f(x) = (ax^2)^2 + 2(ax^2)(b) + b^2$$
$$f(x) = a^2x^4 + 2abx^2 + b^2$$
Now, differentiating both sides with respect to $x$,
$$\frac{d}{dx}\big(f(x)\big) = \frac{d}{dx}(a^2x^4 + 2abx^2 + b^2)$$
Therefore,
$$f'(x) = a^2\frac{d}{dx}(x^4) + 2ab\frac{d}{dx}(x^2) + \frac{d}{dx}(b^2)$$
Using the rule $\dfrac{d}{dx}(x^n) = n x^{n-1}$ and $\dfrac{d}{dx}(\text{constant}) = 0$, we get
$$f'(x) = a^2(4x^3) + 2ab(2x) + 0$$
Simplifying,
$$f'(x) = 4a^2x^3 + 4abx$$
$$
\boxed{f'(x) = 4a^2x^3 + 4abx}
$$
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NCERT Question 7. For some constants $a$ and $b$, find the derivative of $$\dfrac{(x – a)}{(x – b)}$$
Solution:
Let
$$f(x) = \dfrac{x – a}{x – b}$$
We will differentiate this function using the quotient rule, which states that if
$$f(x) = \dfrac{u}{v}, \text{ then } f'(x) = \dfrac{v \dfrac{du}{dx} – u \dfrac{dv}{dx}}{v^2}$$
Here,
$$u = x – a \quad \text{and} \quad v = x – b$$
Now,
$$\dfrac{du}{dx} = 1 \quad \text{and} \quad \dfrac{dv}{dx} = 1$$
Substitute into the quotient rule:
$$f'(x) = \dfrac{(x – b)(1) – (x – a)(1)}{(x – b)^2}$$
Simplify the numerator:
$$f'(x) = \dfrac{(x – b – x + a)}{(x – b)^2}$$
$$f'(x) = \dfrac{(a – b)}{(x – b)^2}$$
$$
\boxed{f'(x) = \dfrac{a – b}{(x – b)^2}}
$$
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NCERT Question 8. Find the derivative of $\dfrac{x^n – a^n}{x – a}$ for some constant $a$.
Solution:
$f(x) = \dfrac{x^n – a^n}{x – a}$
Taking derivative both sides,
$$\frac{d}{dx}(f(x)) = \frac{d}{dx}\left(\frac{x^n – a^n}{x – a}\right)$$
Using quotient rule, we have
$$(\frac{u}{v})’ = \frac{uv’ – vu’}{u^2}$$
$$f'(x)=\left(\frac{(x-a)\frac{d}{dx}(x^n – a^n) – (x^n – a^n)\frac{d}{dx}(x – a)}{(x-a)^2}\right)$$
$$f'(x)=\left(\frac{(x-a)\left[\frac{d}{dx}(x^n) – \frac{d}{dx}(a^n)\right] – (x^n – a^n)(1)}{(x-a)^2}\right)$$
As, the derivative of $x^n$ is $nx^{n-1}$ and derivative of constant is $0$.
$$f'(x)=\left(\frac{(x-a)\left[(nx^{n-1}) – 0)\right] – (x^n – a^n)}{(x-a)^2}\right)$$
$$f'(x)=\left(\frac{(x-a)(nx^{n-1}) – x^n + a^n}{(x-a)^2}\right)$$
$$f'(x)=\left(\frac{(nx^{n-1}+1 – a^n x^{n-1}) – x^n + a^n}{(x-a)^2}\right)$$
$$f'(x)=\left(\frac{(nx^n – a^n x^{n-1}) – x^n + a^n}{(x-a)^2}\right)$$
NCERT Question 9.1 : Find the derivative of $(2x – \dfrac{3}{4})$
Solution:
Let
$$f(x) = 2x – \frac{3}{4}$$
Taking derivative both sides,
$$\frac{d}{dx}\big(f(x)\big) = \frac{d}{dx}\left(2x – \frac{3}{4}\right)$$
We know:
- Derivative of (2x) is (2)
- Derivative of a constant is (0)
Therefore,
$$f'(x) = 2 – 0$$
$$f'(x) = 2$$
NCERT Question 9.2 : Find the derivative of $(5x^3 + 3x – 1)(x – 1)$
Solution :
Let
$$f(x) = (5x^3 + 3x – 1)(x – 1)$$
Using product rule:
$$(uv)’ = u’v + uv’$$
Here,
$$u = 5x^3 + 3x – 1 \quad \text{and} \quad v = x – 1$$
So,
$$u’ = \frac{d}{dx}(5x^3 + 3x – 1) = 15x^2 + 3$$
$$v’ = \frac{d}{dx}(x – 1) = 1$$
Now apply product rule:
$$f'(x) = (15x^2 + 3)(x – 1) + (5x^3 + 3x – 1)(1)$$
Expand step-by-step:
$$(15x^2 + 3)(x – 1) = 15x^3 – 15x^2 + 3x – 3$$
So,
$$f'(x) = 15x^3 – 15x^2 + 3x – 3 + 5x^3 + 3x – 1$$
Combine like terms:
$$f'(x) = (15x^3 + 5x^3) + (-15x^2) + (3x + 3x) + (-3 – 1)$$
$$f'(x) = 20x^3 – 15x^2 + 6x – 4$$
Final Answer
$$\boxed{f'(x) = 20x^3 – 15x^2 + 6x – 4}$$
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NCERT Question 9.3 : Find the derivative of $x^{-3}(5 + 3x)$
Solution :
Let
$$f(x) = x^{-3}(5 + 3x)$$
Apply the product rule:
$$(uv)’ = u’v + uv’$$
Here,
$$u = x^{-3} \quad \Rightarrow \quad u’ = -3x^{-4}$$
$$v = 5 + 3x \quad \Rightarrow \quad v’ = 3$$
Now substitute:
$$f'(x) = (-3x^{-4})(5 + 3x) + (x^{-3})(3)$$
Distribute terms:
$$f'(x) = -15x^{-4} – 9x^{-3} + 3x^{-3}$$
Combine like terms:
$$f'(x) = -15x^{-4} – 6x^{-3}$$
Factor if desired:
$$f'(x) = -3x^{-4}(5 + 2x)$$
Final Answer
$$\boxed{f'(x) = -15x^{-4} – 6x^{-3}}$$
or
$$\boxed{f'(x) = -3x^{-4}(5 + 2x)}$$
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NCERT Question 9.4 : Find the derivative of $x^{5}(3 – 6x^{-9})$
Solution :
Let
$$f(x) = x^{5}(3 – 6x^{-9})$$
Expand first to simplify differentiation:
$$f(x) = 3x^{5} – 6x^{5} \cdot x^{-9}$$
$$f(x) = 3x^{5} – 6x^{-4}$$
Now differentiate term-by-term:
- Derivative of $3x^{5}$:
$$\frac{d}{dx}(3x^5) = 15x^4$$ - Derivative of $-6x^{-4}$:
$$\frac{d}{dx}(-6x^{-4}) = -6(-4)x^{-5} = 24x^{-5}$$
Therefore,
$$f'(x) = 15x^{4} + 24x^{-5}$$
You may also express it in a single fraction:
$$f'(x) = \frac{15x^{9} + 24}{x^{5}}$$
✅ Final Answer
$$\boxed{f'(x) = 15x^{4} + 24x^{-5}}$$
or
$$\boxed{f'(x) = \frac{15x^{9} + 24}{x^{5}}}$$
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NCERT Question 9.5 : Find the derivative of $x^{-4}(3 – 4x^{-5})$ using the product rule.
Solution :
Let
- $u = x^{-4}$
- $v = 3 – 4x^{-5}$
Then,
- $u’ = \frac{d}{dx}(x^{-4}) = -4x^{-5}$
- $v’ = \frac{d}{dx}(3 – 4x^{-5}) = 0 – 4(-5)x^{-6} = 20x^{-6}$
Using the product rule:
$$\frac{d}{dx}(uv) = u’v + uv’$$
Substituting values:
$$f'(x) = (-4x^{-5})(3 – 4x^{-5}) + (x^{-4})(20x^{-6})$$
Now expand and simplify:
First part:
$$(-4x^{-5})(3) = -12x^{-5}$$
$$(-4x^{-5})(-4x^{-5}) = +16x^{-10}$$
Second part:
$$20x^{-4-6} = 20x^{-10}$$
Combine like terms:
$$f'(x) = -12x^{-5} + 16x^{-10} + 20x^{-10}$$
$$f'(x) = -12x^{-5} + 36x^{-10}$$
✅ Final Answer
$$\boxed{f'(x) = -12x^{-5} + 36x^{-10}}$$
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NCERT Question 9.6 : Find the derivative of
$$f(x) = \frac{2}{x+1} – \frac{x^2}{3x – 1}$$
Solution:
Differentiate term-by-term.
For the first term:
$$\frac{2}{x+1} = 2(x+1)^{-1}$$
$$\frac{d}{dx}\left[ 2(x+1)^{-1} \right] = -2(x+1)^{-2}$$
So,
$$\frac{d}{dx}\left[\frac{2}{x+1}\right] = -\frac{2}{(x+1)^2}$$
For the second term, use the quotient rule:
$$\left(\frac{p}{q}\right)’ = \frac{p’q – pq’}{q^2}$$
Here,
$$p = x^2 \Rightarrow p’ = 2x$$
$$q = 3x – 1 \Rightarrow q’ = 3$$
Apply the rule:
$$\frac{d}{dx}\left[\frac{x^2}{3x – 1}\right] = \frac{(2x)(3x – 1) – (x^2)(3)}{(3x – 1)^2}$$
Expand:
$$= \frac{6x^2 – 2x – 3x^2}{(3x – 1)^2}$$
$$= \frac{3x^2 – 2x}{(3x – 1)^2}$$
$$= \frac{x(3x – 2)}{(3x – 1)^2}$$
Final Answer
$$f'(x) = -\frac{2}{(x+1)^2} – \frac{x(3x – 2)}{(3x – 1)^2}$$
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