Anand Classes provides comprehensive and reliable Limits and Derivatives Miscellaneous Exercise NCERT Solutions Class 11 Maths (Set-3) to help students build a strong foundation in calculus. These solutions are prepared according to the latest NCERT and CBSE syllabus, featuring detailed, step-by-step explanations for every problem. Perfect for Class 11 students aiming to improve their conceptual understanding and problem-solving skills, these NCERT solutions make exam preparation simpler and more effective. Click the print button to download study material and notes.
NCERT Question 20 Find the derivative of
$$f(x)=\frac{a+b\sin x}{c+d\cos x}$$
Solution :
Let
$$f(x)=\frac{a+b\sin x}{c+d\cos x}$$
Let $u=a+b\sin x$ and $v=c+d\cos x$.
Then $u’=b\cos x$ and $v’=-d\sin x$.
By the quotient rule,
$$
f'(x)=\frac{u’v-uv’}{v^2}$$
$$f'(x) =\frac{b\cos x(c+d\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}
$$
Simplify the numerator using $\sin^2x+\cos^2x=1$:
$$
\begin{aligned}
\text{numerator}
&=bc\cos x+bd\cos^2x+ad\sin x+bd\sin^2x\\
&=bc\cos x+bd(\cos^2x+\sin^2x)+ad\sin x\\
&=bc\cos x+bd+ad\sin x.
\end{aligned}
$$
Therefore
$$
\boxed{f'(x)=\frac{bc\cos x+bd+ad\sin x}{(c+d\cos x)^2}}
$$
For more worked derivatives and concise notes, check out Anand Classes — great for CBSE & JEE preparation.
NCERT Question 21 : Find the derivative of $$f(x)=\frac{\sin(x+a)}{\cos x}$$
Solution :
Let
$$f(x)=\frac{\sin(x+a)}{\cos x}$$
Using quotient rule:
$$f'(x)=\frac{u’v – uv’}{v^2}$$
Here
$u=\sin(x+a)$
$v=\cos x$
So
$u’=\cos(x+a)$
$v’=-\sin x$
Now substitute:
$$f'(x)=\frac{\cos(x+a)\cos x – \sin(x+a)(- \sin x)}{\cos^2 x}$$
Remove minus sign:
$$f'(x)=\frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}$$
Use identity:
$$\cos A \cos B + \sin A \sin B = \cos(A-B)$$
Take $A=x+a$ and $B=x$:
$$f'(x)=\frac{\cos((x+a)-x)}{\cos^2 x}$$
$$f'(x)=\frac{\cos a}{\cos^2 x}$$
Since
$$\sec x = \frac{1}{\cos x}$$
Final answer:
$$\boxed{f'(x)=\cos a \sec^2 x}$$
NCERT Question 22 : Find the derivative of $$f(x) = x^4 (5\sin x – 3\cos x)$$
Solution :
Let
$$f(x) = x^4 (5\sin x – 3\cos x)$$
Using product rule:
$$f'(x) = u’v + uv’$$
where
$u = x^4$ and $v = 5\sin x – 3\cos x$
So
$$u’ = 4x^3$$
$$v’ = 5\cos x – 3(-\sin x)$$
$$v’ = 5\cos x + 3\sin x$$
Now substitute:
$$f'(x) = x^4(5\cos x + 3\sin x) + (5\sin x – 3\cos x)(4x^3)$$
Factor out $x^3$:
$$f'(x) = x^3\left[x(5\cos x + 3\sin x) + 4(5\sin x – 3\cos x)\right]$$
Expand:
$$f'(x) = x^3\left[5x\cos x + 3x\sin x + 20\sin x – 12\cos x\right]$$
✅ Final simplified expression:
$$\boxed{f'(x) = x^3\left(5x\cos x + 3x\sin x + 20\sin x – 12\cos x\right)}$$
NCERT Question 23 : Find the derivative of $(x^2 + 1)\cos x$
Solution :
Let
$$f(x) = (x^2 + 1)\cos x$$
Using product rule:
$$f'(x) = u’v + uv’$$
where
$u = x^2 + 1$ and $v = \cos x$
So,
$$u’ = 2x$$
$$v’ = -\sin x$$
Substitute:
$$f'(x) = (x^2 + 1)(-\sin x) + \cos x (2x)$$
Expand:
$$f'(x) = -x^2 \sin x – \sin x + 2x \cos x$$
✅ Final Answer:
$$\boxed{f'(x) = -x^2 \sin x – \sin x + 2x \cos x}$$
NCERT Question 24 : Find the derivative of
$$(ax^2 + \sin x)(p + q\cos x)$$
Solution :
Let
$$f(x) = (ax^2 + \sin x)(p + q\cos x)$$
Using the product rule:
$$(uv)’ = u’v + uv’$$
Let
$$u = ax^2 + \sin x \quad \Rightarrow \quad u’ = 2ax + \cos x$$
$$v = p + q\cos x \quad \Rightarrow \quad v’ = -q\sin x$$
Now substitute:
$$f'(x) = (u’v) + (uv’)$$
So,
$$f'(x) = (2ax + \cos x)(p + q\cos x) + (ax^2 + \sin x)(-q\sin x)$$
Rewrite neatly:
$$f'(x) = (p + q\cos x)(2ax + \cos x) – q\sin x(ax^2 + \sin x)$$
✅ Final Answer
$$\boxed{f'(x) = (p + q\cos x)(2ax + \cos x) – q\sin x(ax^2 + \sin x)}$$
NCERT Question 25 : Find the derivative of
$$(x + \cos x)(x – \tan x)$$
Solution :
Let
$$f(x) = (x + \cos x)(x – \tan x)$$
Using the product rule:
$$(uv)’ = u’v + uv’$$
Let
$$u = x + \cos x \quad \Rightarrow \quad u’ = 1 – \sin x$$
$$v = x – \tan x \quad \Rightarrow \quad v’ = 1 – \sec^2 x$$
Now apply the rule:
$$f'(x) = (1 – \sin x)(x – \tan x) + (1 – \sec^2 x)(x + \cos x)$$
✅ Final Answer
$$\boxed{f'(x) = (1 – \sin x)(x – \tan x) + (1 – \sec^2 x)(x + \cos x)}$$
NCERT Question 26 : Find the derivative of
$$\frac{4x + 5\sin x}{3x + 7\cos x}$$
Solution :
Let
$$f(x) = \frac{4x + 5\sin x}{3x + 7\cos x}$$
Using quotient rule:
$$(\frac{u}{v})’ = \frac{vu’ – u,v’}{v^2}$$
Here,
$$u = 4x + 5\sin x \quad \Rightarrow \quad u’ = 4 + 5\cos x$$
$$v = 3x + 7\cos x \quad \Rightarrow \quad v’ = 3 – 7\sin x$$
Substitute into formula:
$$f'(x) = \frac{(3x + 7\cos x)(4 + 5\cos x) – (4x + 5\sin x)(3 – 7\sin x)}{(3x + 7\cos x)^2}$$
✅ Final Answer
$$\boxed{f'(x) = \frac{(3x + 7\cos x)(4 + 5\cos x) – (4x + 5\sin x)(3 – 7\sin x)}{(3x + 7\cos x)^2}}$$
NCERT Question 27 : Find the derivative of $$\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}$$
Solution:
Let
$$f(x)=\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}$$
Taking derivative on both sides,
$$\frac{d}{dx}\big(f(x)\big)=\frac{d}{dx}\left(\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}\right)$$
Using the quotient rule
$$\left(\frac{u}{v}\right)’=\frac{u’v-uv’}{v^2}$$
Here:
$$u=x^2 \cos\left(\frac{\pi}{4}\right)$$
$$v=\sin x$$
Now derivatives:
$$u’=2x \cos\left(\frac{\pi}{4}\right)$$
$$v’=\cos x$$
Substituting:
$$f'(x)=\frac{2x \cos\left(\frac{\pi}{4}\right)\sin x – x^2 \cos\left(\frac{\pi}{4}\right)\cos x}{(\sin x)^2}$$
Factor out $\cos\left(\frac{\pi}{4}\right)$ and $x$:
$$f'(x)=\frac{x \cos\left(\frac{\pi}{4}\right)\left(2\sin x – x\cos x\right)}{\sin^2 x}$$
✅ Final Answer:
$$\boxed{f'(x)=\frac{x \cos\left(\frac{\pi}{4}\right)\left(2\sin x – x\cos x\right)}{\sin^2 x}}$$
NCERT Question 28 : Find the derivative of $$\frac{x}{1 + \tan x}$$
Solution:
Let
$$f(x)=\frac{x}{1+\tan x}$$
Taking derivative on both sides,
$$\frac{d}{dx}\big(f(x)\big)=\frac{d}{dx}\left(\frac{x}{1+\tan x}\right)$$
Using the quotient rule:
$$\left(\frac{u}{v}\right)’=\frac{u’v – uv’}{v^2}$$
Here,
$$u=x \quad , \quad v=1+\tan x$$
$$u’=1 \quad , \quad v’=\frac{d}{dx}(\tan x)$$
Derivative of $\tan x$ using First Principle:
$$g(x)=\tan x=\frac{\sin x}{\cos x}$$
$$g'(x)=\lim_{h\to 0}\frac{\sin(x+h)\cos x – \sin x\cos(x+h)}{h\cos x\cos(x+h)}$$
Using identity
$$\sin a \cos b – \cos a \sin b = \sin(a-b)$$
$$g'(x)=\lim_{h\to 0}\frac{\sin h}{h\cos x\cos(x+h)}$$
Split the limit:
$$g'(x)=\frac{1}{\cos x}\left(\lim_{h\to 0}\frac{1}{\cos(x+h)}\right)\left(\lim_{h\to 0}\frac{\sin h}{h}\right)$$
$$g'(x)=\frac{1}{\cos^2 x}$$
Thus,
$$\frac{d}{dx}(\tan x)=\sec^2 x$$
Substitute back into quotient rule:
$$f'(x)=\frac{(1+\tan x)(1) – x(\sec^2 x)}{(1+\tan x)^2}$$
✅ Final Answer:
$$\boxed{f'(x)=\frac{1+\tan x – x\sec^2 x}{(1+\tan x)^2}}$$
NCERT Question 29 : Find the derivative of $$(x + \sec x)(x – \tan x)$$
Solution:
Let
$$f(x) = (x + \sec x)(x – \tan x)$$
Taking derivative on both sides,
$$\frac{d}{dx}\big(f(x)\big)=\frac{d}{dx}\left((x+\sec x)(x-\tan x)\right)$$
Using the product rule:
$$(uv)’ = u’v + uv’$$
Here,
$$u = x + \sec x,\quad v = x – \tan x$$
$$u’ = 1 + \frac{d}{dx}(\sec x),\quad v’ = 1 – \frac{d}{dx}(\tan x)$$
From first principle (already shown earlier):
$$\frac{d}{dx}(\tan x) = \sec^2 x$$
$$\frac{d}{dx}(\sec x) = \sec x\tan x$$
Substitute into derivative:
$$f'(x) = (x+\sec x)\big(1 – \sec^2 x\big) + (x-\tan x)\big(1 + \sec x\tan x\big)$$
Expand each term separately:
First term:
$$ (x+\sec x)(1 – \sec^2 x) = x(1 – \sec^2 x) + \sec x(1 – \sec^2 x) $$
Second term:
$$ (x-\tan x)(1 + \sec x\tan x) = x(1 + \sec x\tan x) – \tan x(1 + \sec x\tan x) $$
✅ Final Answer (simplified form):
$$\boxed{f'(x) = (x+\sec x)(1 – \sec^2 x) + (x-\tan x)(1 + \sec x\tan x)}$$
NCERT Question 30 : Find the derivative of $$\frac{x}{\sin^n x}$$
Solution:
Let
$$f(x) = \frac{x}{\sin^n x}$$
Taking derivative both sides,
$$\frac{d}{dx}\big(f(x)\big) = \frac{d}{dx}\left(\frac{x}{\sin^n x}\right)$$
Using the quotient rule:
$$(u/v)’ = \frac{uv’ – u’v}{v^2}$$
So,
$$f'(x) = \dfrac{(\sin^n x)\dfrac{d}{dx}(x) – x\dfrac{d}{dx}(\sin^n x)}{(\sin^n x)^2}$$
As the derivative of $x$ is $1$,
$$f'(x) = \dfrac{\sin^n x – x\dfrac{d}{dx}(\sin^n x)}{(\sin^n x)^2}$$
Let $g(x) = \sin^n x$
We know from the general rule (proved using product rule & induction):
$$\frac{d}{dx}(\sin^n x) = n \sin^{,n-1}x\cos x$$
So substituting:
$$f'(x) = \dfrac{\sin^n x – x\big(n \sin^{,n-1}x\cos x\big)}{(\sin^n x)^2}$$
✅ Final Answer:
$$\boxed{f'(x) = \dfrac{\sin^n x – xn\sin^{,n-1}x\cos x}{\sin^{2n}x}}$$
