Anand Classes presents well-structured and easy-to-understand NCERT Solutions for Limits and Derivatives Miscellaneous Exercise Class 11 (Set-2) designed to help students master calculus concepts effectively. These solutions are prepared as per the latest NCERT and CBSE guidelines, providing step-by-step explanations for every question to ensure complete conceptual clarity. Ideal for Class 11 students aiming for academic excellence, these solutions simplify complex problems and make revision effortless. Click the print button to download study material and notes.
NCERT Question 11 : Find the derivative of
$$f(x) = 4\sqrt{x} – 2$$
Solution :
Rewrite $\sqrt{x}$ using exponent form:
$$f(x) = 4x^{\frac{1}{2}} – 2$$
Differentiate both sides:
$$f'(x) = \frac{d}{dx}\left(4x^{\frac{1}{2}}\right) – \frac{d}{dx}(2)$$
Using $\frac{d}{dx}(x^n) = nx^{n-1}$ and derivative of constant is $0$:
$$f'(x) = 4 \cdot \frac{1}{2} x^{\frac{1}{2}-1} – 0$$
$$f'(x) = 2x^{-\frac{1}{2}}$$
Rewrite using radicals:
$$f'(x) = \frac{2}{\sqrt{x}}$$
Final Answer
$$\boxed{f'(x) = \frac{2}{\sqrt{x}}}$$
NCERT Question 12 : Find the derivative of
$$f(x) = (ax + b)^n$$
using first principle.
Solution :
Let
$$f(x) = (ax + b)^n$$
Then,
$$f(x+h) = (a(x+h) + b)^n = (ax + ah + b)^n$$
Using the first principle of differentiation:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
Substitute values:
$$f'(x) = \lim_{h \to 0} \frac{(ax + ah + b)^n – (ax + b)^n}{h}$$
Factor out $(ax+b)^n$:
$$f'(x) = (ax+b)^n \lim_{h \to 0} \frac{\left(1 + \frac{ah}{ax+b}\right)^n – 1}{h}$$
Using the Binomial Expansion:
$$\left(1 + \frac{ah}{ax+b}\right)^n = 1 + n\frac{ah}{ax+b} + \frac{n(n-1)}{2!}\left(\frac{ah}{ax+b}\right)^2 + \cdots$$
Substitute:
$$f'(x) = (ax+b)^n \lim_{h \to 0} \frac{n\frac{ah}{ax+b} + \frac{n(n-1)}{2!}\left(\frac{ah}{ax+b}\right)^2 + \cdots}{h}$$
Factor $h$ out:
$$f'(x) = (ax+b)^n \lim_{h \to 0} \left[ n\frac{a}{ax+b} + \frac{n(n-1)a^2h}{2!(ax+b)^2} + \cdots \right]$$
As $h \to 0$, all higher terms vanish:
$$f'(x) = (ax+b)^n \left( n\frac{a}{ax+b} \right)$$
Simplifying:
$$f'(x) = na(ax+b)^{n-1}$$
Final Answer
$$\boxed{f'(x) = na(ax + b)^{,n-1}}$$
NCERT Question 13 : Find the derivative of
$$f(x)=(ax+b)^n (cx+d)^m$$
using first principle.
Solution :
Let
$$f(x) = (ax+b)^n (cx+d)^m$$
By first principle:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
Now,
$$f(x+h) = (a(x+h)+b)^n (c(x+h)+d)^m $$
$$f(x+h) = (ax+ah+b)^n (cx+ch+d)^m$$
So,
$$f(x+h)-f(x) = (ax+ah+b)^n (cx+ch+d)^m – (ax+b)^n (cx+d)^m$$
Add and subtract $(ax+ah+b)^n (cx+d)^m$ in above equation
inside:
$f(x+h)-f(x) = (ax+ah+b)^n (cx+d)^m – (ax+b)^n (cx+d)^m + (ax+ah+b)^n (cx+ch+d)^m – (ax+ah+b)^n (cx+d)^m$
Divide by (h):
$$
\frac{f(x+h)-f(x)}{h} =
\frac{(ax+ah+b)^n – (ax+b)^n}{h}(cx+d)^m +\\
(ax+ah+b)^n \frac{(cx+ch+d)^m – (cx+d)^m}{h}
$$
Take the limit $(h \to 0):$
- Binomial expansion gives
$$\lim_{h\to0}\frac{(ax+ah+b)^n – (ax+b)^n}{h} = na(ax+b)^{n-1}$$ - And similarly
$$\lim_{h\to0}\frac{(cx+ch+d)^m – (cx+d)^m}{h} = mc(cx+d)^{m-1}$$
Substitute the limits:
$$f'(x) = na(ax+b)^{n-1}(cx+d)^m + mc(ax+b)^n(cx+d)^{m-1}$$
✅ Final Answer
$$
\boxed{
f'(x) = na(ax+b)^{n-1}(cx+d)^m + mc(ax+b)^n(cx+d)^{m-1}
}
$$
NCERT Question 14 : Find the derivative of $\sin(x+a)$ using first principle
Solution:
Let
$$f(x) = \sin(x+a)$$
Then
$$f(x+h) = \sin((x+h) + a) = \sin(x + a + h)$$
Using first principle:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
Substituting the values:
$$f'(x) = \lim_{h \to 0} \frac{\sin(x+a+h) – \sin(x+a)}{h}$$
Using identity:
$$\sin A – \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
Let
$(A = x+a+h)$ and $(B = x+a)$
So:
$$
\sin(x+a+h) – \sin(x+a)
= 2 \cos\left(\frac{2(x+a)+h}{2}\right)\sin\left(\frac{h}{2}\right)
$$
Substitute into limit:
$$
f'(x) = \lim_{h \to 0}
\frac{2 \cos\left(x+a+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}
$$
Rewrite:
$$
f'(x) = \lim_{h \to 0}
\left[2 \cos\left(x+a+\frac{h}{2}\right)\cdot
\frac{\sin\left(\frac{h}{2}\right)}{h}\right]
$$
Multiply and divide by 2:
$$
f'(x) =
\lim_{h \to 0}
\left[
\cos\left(x+a+\frac{h}{2}\right)
\cdot
\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}
\right]
$$
Now using the limit identity
$$\lim_{t\to0} \frac{\sin t}{t} = 1$$
As $(h \to 0),$
$$\cos\left(x+a+\frac{h}{2}\right) \to \cos(x+a)$$
Thus,
$$
\boxed{f'(x) = \cos(x+a)}
$$
NCERT Question 15: Find the derivative of $\csc x \cot x$
Solution:
Let
$$f(x) = \csc x \cot x$$
Taking derivative both sides,
$$\frac{d}{dx}(f(x)) = \frac{d}{dx}(\csc x \cot x)$$
Using the product rule:
$$(uv)’ = uv’ + u’v$$
So,
$$f'(x) = \cot x \cdot \frac{d}{dx}(\csc x) + (\csc x)\cdot \frac{d}{dx}(\cot x)$$
Now using standard derivatives:
$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$
$$\frac{d}{dx}(\cot x) = -\csc^2 x$$
Therefore,
$$f'(x) = \cot x (-\csc x \cot x) + (\csc x)(-\csc^2 x)$$
$$f'(x) = -\cot^2 x \csc x – \csc^3 x$$
NCERT Question 16: Find the derivative of $$\frac{\cos x}{1+\sin x}$$
Solution:
Let
$$f(x) = \frac{\cos x}{1 + \sin x}$$
Taking derivative both sides:
$$\frac{d}{dx}(f(x)) = \frac{d}{dx}\left(\frac{\cos x}{1+\sin x}\right)$$
Using the quotient rule:
$$(\frac{u}{v})’ = \frac{v u’ – u v’}{v^2}$$
So,
$$f'(x) = \frac{(1+\sin x)\frac{d}{dx}(\cos x) – (\cos x)\frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}$$
$$f'(x) = \frac{(1+\sin x)(-\sin x) – (\cos x)(\cos x)}{(1+\sin x)^2}$$
$$f'(x) = \frac{-\sin x – \sin^2 x – \cos^2 x}{(1+\sin x)^2}$$
Using identity:
$$\sin^2 x + \cos^2 x = 1$$
$$f'(x) = \frac{-\sin x – 1}{(1+\sin x)^2}$$
$$f'(x) = -\frac{(\sin x + 1)}{(1+\sin x)^2}$$
Cancel common term:
$$f'(x) = -\frac{1}{1 + \sin x}$$
NCERT Question 17: Find the derivative of $$f(x) = \frac{\sin x + \cos x}{\sin x – \cos x}$$
Solution:
Let
$$f(x) = \frac{\sin x + \cos x}{\sin x – \cos x}$$
Taking derivative both sides:
$$f'(x) = \frac{d}{dx}\left(\frac{\sin x + \cos x}{\sin x – \cos x}\right)$$
Using quotient rule:
$$\left(\frac{u}{v}\right)’ = \frac{v u’ – u v’}{v^2}$$
Here,
$(u = \sin x + \cos x)$ and $(v = \sin x – \cos x)$
So:
$$u’ = \cos x – \sin x$$
$$v’ = \cos x + \sin x$$
Substitute:
$$f'(x) = \frac{(\sin x – \cos x)(\cos x – \sin x) – (\sin x + \cos x)(\cos x + \sin x)}{(\sin x – \cos x)^2}$$
Expand:
$$f'(x) = \frac{-(\sin x – \cos x)^2 – (\sin x + \cos x)^2}{(\sin x – \cos x)^2}$$
Use identities:
$$(\sin x – \cos x)^2 = \sin^2 x – 2\sin x\cos x + \cos^2 x = 1 – 2\sin x\cos x$$
$$(\sin x + \cos x)^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + 2\sin x\cos x$$
So:
$$f'(x) = \frac{-(1 – 2\sin x\cos x) – (1 + 2\sin x\cos x)}{(\sin x – \cos x)^2}$$
$$f'(x) = \frac{-1 + 2\sin x\cos x -1 – 2\sin x\cos x}{(\sin x – \cos x)^2}$$
$$f'(x) = \frac{-2}{(\sin x – \cos x)^2}$$
Final Answer:
$$\boxed{f'(x) = -\frac{2}{(\sin x – \cos x)^2}}$$
NCERT Question 18 Find the derivative of
$$f(x)=\dfrac{\sec x – 1}{\sec x + 1}$$
Solution :
Given,
$$f(x)=\dfrac{\sec x – 1}{\sec x + 1}$$
Rewrite using $\sec x = \frac{1}{\cos x}$:
$$f(x)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}
=\dfrac{\dfrac{1-\cos x}{\cos x}}{\dfrac{1+\cos x}{\cos x}}
=\dfrac{1-\cos x}{1+\cos x}$$
Now differentiate:
$$f'(x)=\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)$$
Let
$u = 1 – \cos x \Rightarrow u’ = \sin x$
$v = 1 + \cos x \Rightarrow v’ = -\sin x$
Using quotient rule:
$$f'(x)=\frac{u’v – uv’}{v^2}$$
Substitute values:
$$f'(x)=\frac{\sin x(1+\cos x)-(1-\cos x)(-\sin x)}{(1+\cos x)^2}$$
Simplify numerator:
$$f'(x) =\frac{\sin x + \sin x\cos x + \sin x – \sin x\cos x}{(1+\cos x)^2}$$
$$f'(x) =\frac{2\sin x}{(1+\cos x)^2}$$
✅ Final Answer
$$\boxed{f'(x)=\frac{2\sin x}{(1+\cos x)^2}}$$
NCERT Question 19 Find the derivative of
$$f(x)=\sin^n x$$
Solution :
For integer $n\ge1$ consider
$$f(x)=\sin^n x=(\sin x)^n.$$
Using the chain rule (or by induction/product-rule as you started), differentiate:
$$
\frac{d}{dx}\big(\sin^n x\big)
= n(\sin x)^{,n-1}\cdot\frac{d}{dx}(\sin x)
= n\sin^{,n-1}x\cos x.
$$
Check small $n$:
$$n=1:\quad \frac{d}{dx}(\sin x)=\cos x.$$
$$n=2:\quad \frac{d}{dx}(\sin^2 x)=2\sin x\cos x.$$
$$n=3:\quad \frac{d}{dx}(\sin^3 x)=3\sin^2 x\cos x.$$
Thus for integer $n$,
$$\boxed{\dfrac{d}{dx}\big(\sin^n x\big)=n\sin^{,n-1}x\cos x.}$$
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