Anand Classes provides detailed and accurate Limits and Derivatives NCERT Solutions Miscellaneous Exercise Class 11 (Set-1) to help students strengthen their understanding of calculus concepts. These solutions follow the latest NCERT and CBSE guidelines, offering clear, step-by-step explanations for each question. Perfect for revision and exam preparation, these Class 11 Maths solutions make learning Limits and Derivatives easier and more effective for students aiming to score high marks. Click the print button to download study material and notes.
Question 1(i) : Find the derivative of $-x$ from first principle.
Solution :
Let $f(x)=-x$.
Then $f(x+h)=-(x+h)$.
From first principle,
$$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{-(x+h)-(-x)}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{-x-h+x}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{-h}{h}.$$
Hence
$$
f'(x)=\lim_{h\to 0}-1=-1.
$$
$$\boxed{f'(x)=-1,}$$
Question 1(ii) : Find the derivative of $(-x)^{-1}$ from first principle.
Solution :
Let $f(x)=(-x)^{-1}=\dfrac{1}{-x}=-\dfrac{1}{x}$.
Then
$$f(x+h)=-\frac{1}{x+h}.$$
From first principle,
$$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{-\dfrac{1}{x+h}-\big(-\dfrac{1}{x}\big)}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{-\dfrac{1}{x+h}+\dfrac{1}{x}}{h}$$
Combine the numerator:
$$
-\frac{1}{x+h}+\frac{1}{x}
=\frac{-x + (x+h)}{x(x+h)}
=\frac{h}{x(x+h)}.
$$
Thus
$$
f'(x)=\lim_{h\to 0}\frac{\dfrac{h}{x(x+h)}}{h}
=\lim_{h\to 0}\frac{1}{x(x+h)}
=\frac{1}{x^2}.
$$
$$\boxed{f'(x)=\frac{1}{x^2}}$$
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Question 1(iii) : Find the derivative of $\sin(x+1)$ from first principle.
Solution:
Let
$$f(x)=\sin(x+1)$$
Then
$$f(x+h)=\sin((x+h)+1)=\sin(x+h+1)$$
From first principle,
$$
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
=\lim_{h\to 0}\frac{\sin(x+h+1)-\sin(x+1)}{h}
$$
Using identity:
$$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
Here,
$A=x+h+1$ and $B=x+1$
So,
$$
f'(x)=\lim_{h\to 0}
\frac{2\cos\left(\dfrac{(x+h+1)+(x+1)}{2}\right)\sin\left(\dfrac{(x+h+1)-(x+1)}{2}\right)}{h}
$$
Simplify inside:
$$
f'(x)=\lim_{h\to 0}
\frac{2\cos\left(\dfrac{2x+h+2}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}
$$
$$
=\lim_{h\to 0}
2\cos(x+1+\tfrac{h}{2})\cdot \frac{\sin(h/2)}{h}
$$
Multiply and divide by 2:
$$
f'(x)=\lim_{h\to 0}
2\cos(x+1+\tfrac{h}{2})\cdot
\frac{\sin(h/2)}{(h/2)}\cdot \frac{1}{2}
$$
Now apply limits:
$\lim\limits_{h\to 0}\cos(x+1+\frac{h}{2})=\cos(x+1)$
$\lim\limits_{h\to 0}\dfrac{\sin(h/2)}{(h/2)}=1$
Thus,
$$
f'(x)=\cos(x+1)
$$
✅ Final Answer:
$$\boxed{f'(x)=\cos(x+1)}$$
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Question 1(iv) : Find the derivative of $\cos\left(x – \frac{\pi}{8}\right)$ from first principle.
Solution:
Let
$$f(x)=\cos\left(x-\frac{\pi}{8}\right)$$
Then
$$f(x+h)=\cos\left((x+h)-\frac{\pi}{8}\right)$$
Using first principle:
$$
f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
=\lim_{h\to0}\frac{\cos\left(x+h-\frac{\pi}{8}\right)-\cos\left(x-\frac{\pi}{8}\right)}{h}
$$
Use identity:
$$\cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
Here,
$A=x+h-\frac{\pi}{8}$ and $B=x-\frac{\pi}{8}$
So,
$$
f'(x)=\lim_{h\to0}
\frac{-2\sin\left(\frac{2x+h-\frac{\pi}{4}}{2}\right)\sin\left(\frac{h}{2}\right)}{h}
$$
Simplify:
$$
f'(x)=\lim_{h\to0}
-2\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h}
$$
Multiply and divide by 2:
$$
f'(x)=\lim_{h\to0}
-2\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\cdot
\frac{\sin(h/2)}{(h/2)}\cdot \frac{1}{2}
$$
Apply limits:
$\lim_{h\to0}\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)=\sin\left(x-\frac{\pi}{8}\right)$
$\lim_{h\to0}\dfrac{\sin(h/2)}{(h/2)}=1$
Thus,
$$
f'(x)=-\sin\left(x-\frac{\pi}{8}\right)
$$
✅ Final Answer:
$$\boxed{f'(x)=-\sin\left(x-\frac{\pi}{8}\right)}$$
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):
NCERT Question 2: Find the derivative of $(x+a)$
Solution:
Let
$$f(x)=x+a$$
Taking derivative of both sides:
$$\frac{d}{dx}(f(x))=\frac{d}{dx}(x+a)$$
We know:
- Derivative of $x$ is $1$
- Derivative of a constant $a$ is $0$
Thus,
$$f'(x)=1+0=1$$
✅ Final Answer:
$$\boxed{f'(x)=1}$$
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NCERT Question 3: Find the derivative of $$(px+q)\left(\frac{r}{x}+s\right)$$
Solution:
Let
$$f(x)=(px+q)\left(\frac{r}{x}+s\right)$$
Use the product rule:
If $f(x)=u \cdot v$ then
$$f'(x)=u’v + uv’$$
Here,
$$u = px+q \quad \Rightarrow \quad u’ = p$$
$$v = \frac{r}{x}+s = r x^{-1} + s \quad \Rightarrow \quad v’ = -r x^{-2} + 0 = -\frac{r}{x^2}$$
Now apply the product rule:
$$f'(x)=p\left(\frac{r}{x}+s\right)+(px+q)\left(-\frac{r}{x^2}\right)$$
So the final answer is:
$$\boxed{f'(x)=p\left(\frac{r}{x}+s\right)-\frac{r(px+q)}{x^2}}$$
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NCERT Question 4: Find the derivative of $(ax+b)(cx+d)^2$
Solution :
Let
$$f(x) = (ax+b)(cx+d)^2$$
Taking derivative of both sides:
$$\frac{d}{dx}[f(x)] = \frac{d}{dx}[(ax+b)(cx+d)^2]$$
Using the product rule $(uv)’ = uv’ + u’v$:
Let
$$u = ax+b \quad \Rightarrow \quad u’ = a$$
$$v = (cx+d)^2 \quad \Rightarrow \quad v’ = 2(cx+d) \cdot c = 2c(cx+d)$$
(using chain rule)
Therefore,
$$f'(x) = (ax+b)\cdot 2c(cx+d) + (cx+d)^2 \cdot a$$
So the final derivative is:
$$\boxed{f'(x) = 2c(ax+b)(cx+d) + a(cx+d)^2}$$
You may also factor if required:
$$\boxed{f'(x) = (cx+d)\left[2c(ax+b) + a(cx+d)\right]}$$
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NCERT Question 5: Find the derivative of $\dfrac{ax+b}{cx+d}$
Solution :
Let
$$f(x) = \frac{ax+b}{cx+d}$$
Taking derivative of both sides:
$$\frac{d}{dx}[f(x)] = \frac{d}{dx}\left(\frac{ax+b}{cx+d}\right)$$
Using the quotient rule:
$$(\frac{u}{v})’ = \frac{uv’ – u’v}{v^2}$$
Here,
$$u = ax+b \quad \Rightarrow \quad u’ = a$$
$$v = cx+d \quad \Rightarrow \quad v’ = c$$
Apply the rule:
$$f'(x) = \frac{(ax+b)'(cx+d) – (ax+b)(cx+d)’}{(cx+d)^2}$$
$$f'(x) = \frac{a(cx+d) – (ax+b)c}{(cx+d)^2}$$
Expand the numerator:
$$f'(x) = \frac{acx + ad – acx – bc}{(cx+d)^2}$$
Simplify:
$$\boxed{f'(x) = \frac{ad – bc}{(cx+d)^2}}$$
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NCERT Question 6 : Find the derivative of $$f(x) = \dfrac{1 + \dfrac{1}{x}}{1 – \dfrac{1}{x}}$$
Solution :
Given:
$$f(x) = \dfrac{1 + \dfrac{1}{x}}{1 – \dfrac{1}{x}}$$
Rewrite by taking LCM in numerator and denominator:
$$f(x) = \dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}$$
Cancel common denominator $x$:
$$f(x) = \dfrac{x+1}{x-1}$$
Now differentiate using quotient rule:
Let,
$$u = x + 1 \Rightarrow u’ = 1$$
$$v = x – 1 \Rightarrow v’ = 1$$
Quotient rule:
$$(\dfrac{u}{v})’ = \dfrac{vu’ – uv’}{v^2}$$
Apply:
$$f'(x) = \dfrac{(x-1)(1) – (x+1)(1)}{(x-1)^2}$$
Simplify numerator:
$$f'(x) = \frac{x-1 – x – 1}{(x-1)^2}$$
$$f'(x) = \frac{-2}{(x-1)^2}$$
✅ Final Answer
$$\boxed{f'(x) = \frac{-2}{(x-1)^2}}$$
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NCERT Question 7 : Find the derivative of:
$$f(x) = \frac{1}{ax^2 + bx + c}$$
Solution :
Given:
$$f(x) = \frac{1}{ax^2 + bx + c}$$
Rewrite in power form:
$$f(x) = (ax^2 + bx + c)^{-1}$$
Differentiate using chain rule:
Let
$$u = ax^2 + bx + c \quad \Rightarrow \quad u’ = 2ax + b$$
So,
$$f'(x) = -1 \cdot (ax^2 + bx + c)^{-2} \cdot (2ax + b)$$
Thus,
$$f'(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$$
✅ Final Answer
$$\boxed{f'(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}}$$
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NCERT Question 8 : Find the derivative of
$$f(x) = \frac{ax + b}{px^2 + qx + r}$$
Solution :
Given,
$$f(x) = \frac{ax + b}{px^2 + qx + r}$$
Using the quotient rule:
$$\left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2}$$
Let
$u = ax + b \Rightarrow u’ = a$
$v = px^2 + qx + r \Rightarrow v’ = 2px + q$
So,
$$f'(x) = \frac{a(px^2 + qx + r) – (ax + b)(2px + q)}{(px^2 + qx + r)^2}$$
Expanding the numerator:
$$
a(px^2 + qx + r) = apx^2 + aqx + ar
$$
$$
(ax + b)(2px + q) = 2apx^2 + aqx + 2bpx + bq
$$
Subtract:
$$
apx^2 + aqx + ar – (2apx^2 + aqx + 2bpx + bq)
$$
Simplifying:
$$
= -apx^2 – 2bpx + ar – bq
$$
✅ Final Answer
$$\boxed{f'(x) = \frac{-apx^2 – 2bpx + ar – bq}{(px^2 + qx + r)^2}}$$
NCERT Question 9 : Find the derivative of
$$f(x) = \frac{px^2 + qx + r}{ax + b}$$
Solution :
Given,
$$f(x) = \frac{px^2 + qx + r}{ax + b}$$
Using the quotient rule:
$$\left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2}$$
Let
$$
u = px^2 + qx + r \quad \Rightarrow \quad u’ = 2px + q
$$
$$
v = ax + b \quad \Rightarrow \quad v’ = a
$$
$$
f'(x) = \frac{(2px + q)(ax + b) – (px^2 + qx + r)(a)}{(ax + b)^2}
$$
Expand the numerator:
$$
(2px + q)(ax + b) = 2apx^2 + aqx + 2bpx + bq
$$
$$
a(px^2 + qx + r) = apx^2 + aqx + ar
$$
Subtract:
$$
2apx^2 + aqx + 2bpx + bq – (apx^2 + aqx + ar)
$$
Simplify:
$$
= apx^2 + 2bpx + bq – ar
$$
✅ Final Answer
$$
\boxed{f'(x) = \frac{apx^2 + 2bpx + bq – ar}{(ax + b)^2}}
$$
NCERT Question 10 Find the derivative of
$$f(x) = \frac{a}{x^4} – \frac{b}{x^2} + \cos x$$
Solution :
Given,
$$f(x) = \frac{a}{x^4} – \frac{b}{x^2} + \cos x$$
Rewrite using negative exponents:
$$f(x) = a x^{-4} – b x^{-2} + \cos x$$
Differentiate both sides:
$$f'(x) = \frac{d}{dx}(a x^{-4}) – \frac{d}{dx}(b x^{-2}) + \frac{d}{dx}(\cos x)$$
Using $\frac{d}{dx}(x^n) = n x^{n-1}$ and $\frac{d}{dx}(\cos x) = -\sin x$:
$$f'(x) = a(-4x^{-5}) – b(-2x^{-3}) – \sin x$$
Simplify:
$$f'(x) = -4ax^{-5} + 2bx^{-3} – \sin x$$
Write back in fractional form:
$$f'(x) = -\frac{4a}{x^5} + \frac{2b}{x^3} – \sin x$$
Final Answer
$$\boxed{f'(x) = -\frac{4a}{x^5} + \frac{2b}{x^3} – \sin x}$$
