Anand Classes offers expertly crafted Limits and Derivatives NCERT Solutions Exercise 12.2 Class 11 (Set-3) of Chapter-12 to help students gain a deep understanding of calculus concepts in a simple and structured way. These solutions strictly follow the latest CBSE syllabus and NCERT textbook, providing clear, step-by-step explanations for every question. Ideal for Class 11 students preparing for exams, these solutions make complex topics easier to grasp and apply. Click the print button to download study material and notes.
NCERT Question 10 : Find the derivative of $\cos x$ from first principle.
Solution:
Given,
$$f(x) = \cos x$$
So,
$$f(x+h) = \cos(x+h)$$
Using first principle:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
Substitute values:
$$f'(x) = \lim_{h \to 0} \frac{\cos(x+h) – \cos x}{h}$$
Using the trigonometric identity:
$$\cos A – \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
Let $A = x+h$ and $B = x$
$$\cos(x+h) – \cos x = -2 \sin\left(\frac{(x+h)+x}{2}\right)\sin\left(\frac{(x+h)-x}{2}\right)$$
$$ = -2 \sin\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)$$
So,
$$f'(x) = \lim_{h \to 0} \dfrac{-2 \sin\left(\dfrac{2x+h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}$$
Now multiply and divide by 2:
$$f'(x) = \lim_{h \to 0} \left[-2 \sin\left(\dfrac{2x+h}{2}\right)\right] \cdot \lim_{h \to 0} \dfrac{\sin\left(\dfrac{h}{2}\right)}{h} \cdot 2$$
Rewrite:
$$f'(x) = \lim_{h \to 0} \left[-\sin\left(\dfrac{2x+h}{2}\right)\right] \cdot \lim_{h \to 0} \dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}$$
We know:
$$\lim_{y \to 0} \frac{\sin y}{y} = 1$$
Thus:
$$f'(x) = -\sin\left(\frac{2x+0}{2}\right) \cdot 1$$
$$f'(x) = -\sin x$$
Final Answer
$$\boxed{f'(x) = -\sin x}$$
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NCERT Question 11(i). Find the derivative of $ \sin x\cos x $ from first principle
Solution:
Let
$$f(x)=\sin x\cos x.$$
Then
$$f(x+h)=\sin(x+h)\cos(x+h).$$
From first principle,
$$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
$$ f'(x) =\lim_{h\to 0}\frac{\sin(x+h)\cos(x+h)-\sin x\cos x}{h}.$$
Use the product-to-sum identity $$ \sin A\cos B=\tfrac{1}{2}\big(\sin(A+B)+\sin(A-B)\big) $$
Applying it to both terms,
$$ \sin(x+h)\cos(x+h)=\tfrac{1}{2}\sin(2x+2h)$$
$$\sin x\cos x=\tfrac{1}{2}\sin(2x).$$
So
$$ f'(x)=\lim_{h\to 0}\frac{\tfrac{1}{2}\sin(2x+2h)-\tfrac{1}{2}\sin(2x)}{h}$$
$$f'(x)=\tfrac{1}{2}\lim_{h\to 0}\frac{\sin(2x+2h)-\sin(2x)}{h}.$$
Use the identity $$\sin A-\sin B=2\cos\big(\dfrac{A+B}{2}\big)\sin\big(\dfrac{A-B}{2}\big)$$ with $A=2x+2h$ and $B=2x$:
$$
\sin(2x+2h)-\sin(2x)=2\cos(2x+h)\sin(h).
$$
Thus
$$ f'(x)=\tfrac{1}{2}\lim_{h\to 0}\frac{2\cos(2x+h)\sin(h)}{h}$$
$$ f'(x) =\lim_{h\to 0}\cos(2x+h)\cdot\frac{\sin(h)}{h}$$
Since $\displaystyle\lim_{h\to 0}\frac{\sin(h)}{h}=1$ and $\displaystyle\lim_{h\to 0}\cos(2x+h)=\cos(2x)$, we get
$$
f'(x)=\cos(2x)\cdot 1=\cos(2x).
$$
$$\boxed{\dfrac{d}{dx}\big(\sin x\cos x\big)=\cos(2x)}$$
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NCERT Question 11 (ii) : Find the derivative of sec x from first principle.
Solution:
Let
$ f(x) = \sec(x) $
Then,
$ f(x+h) = \sec(x+h) $
Using the first principle of derivative:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}
$$
$$
f'(x) = \lim_{h \to 0} \frac{\sec(x+h) – \sec(x)}{h}
$$
Rewrite using $\sec(x) = \dfrac{1}{\cos(x)}$:
$$
f'(x) = \lim_{h \to 0} \dfrac{\dfrac{1}{\cos(x+h)} – \dfrac{1}{\cos(x)}}{h}
$$
Take LCM:
$$
f'(x) = \lim_{h \to 0} \frac{\cos(x) – \cos(x+h)}{h \cos(x+h)\cos(x)}
$$
Use identity:
$\cos A – \cos B = -2 \sin\Big(\frac{A+B}{2}\Big)\sin\Big(\frac{A-B}{2}\Big)$
Let $A = x+h$ and $B = x$:
$$
\cos(x) – \cos(x+h)
= 2 \sin\Big(\frac{2x+h}{2}\Big)\sin\Big(\frac{h}{2}\Big)
$$
So:
$$
f'(x) =
\lim_{h \to 0}
\frac{2 \sin\left(\dfrac{2x+h}{2}\right)\sin\left(\dfrac{h}{2}\right)}
{h \cos(x+h)\cos(x)}
$$
Separate limits:
$$
f'(x) =
\lim_{h \to 0}
\frac{2\sin\left(x + \dfrac{h}{2}\right)}{\cos(x+h)\cos(x)}
\cdot
\lim_{h \to 0}
\frac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}
$$
We know:
$$
\lim_{h \to 0} \dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} = 1
$$
Now take the limit:
$$
f'(x) =
\dfrac{2 \sin(x)}{\cos^2(x)}
$$
Use
$\dfrac{\sin(x)}{\cos(x)} = \tan(x)$
$\dfrac{1}{\cos(x)} = \sec(x)$
$$
f'(x) = \sec(x)\tan(x)
$$
✅ Final Answer
$$
\boxed{f'(x) = \sec(x)\tan(x)}
$$
NCERT Question 11 (iii) : Find the derivative of $5\sec x + 4\cos x$
Solution:
Let
$$f(x) = 5\sec x + 4\cos x$$
Differentiate both sides:
$$
f'(x) = \frac{d}{dx}(5\sec x) + \frac{d}{dx}(4\cos x)
$$
Using standard derivatives:
$\frac{d}{dx}(\sec x) = \sec x \tan x$
$\frac{d}{dx}(\cos x) = -\sin x$
So,
$$
f'(x) = 5(\sec x \tan x) + 4(-\sin x)
$$
$$
f'(x) = 5\sec x \tan x – 4\sin x
$$
✅ Final Answer
$$
\boxed{f'(x) = 5\sec x \tan x – 4\sin x}
$$
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Question 11 (iv). Find the derivative of cosec x from first principle
Solution:
Let
$$f(x)=\csc x=\frac{1}{\sin x}.$$
Using the first principle,
$$
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
=\lim_{h\to 0}\frac{\csc(x+h)-\csc x}{h}.
$$
Write in sine form and combine:
$$
f'(x) =\lim_{h\to 0}\dfrac{\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin x}}{h}$$
$$f'(x) =\lim_{h\to 0}\dfrac{\sin x-\sin(x+h)}{h\sin x\sin(x+h)}.$$
Use the identity $$\sin A-\sin B=2\cos\big(\frac{A+B}{2}\big)\sin\big(\frac{A-B}{2}\big)$$ with $A=x$ and $B=x+h$:
$$
\sin x-\sin(x+h)= -2\cos\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right).
$$
Substitute:
$$
f'(x)=\lim_{h\to 0}\frac{-2\cos\big(x+\frac{h}{2}\big)\sin\big(\frac{h}{2}\big)}
{h\sin x\sin(x+h)}.
$$
Rearrange factors:
$$
f'(x)= -\lim_{h\to 0}\frac{\cos\big(x+\frac{h}{2}\big)}{\sin x\sin(x+h)}
\cdot\frac{2\sin\big(\frac{h}{2}\big)}{h}.
$$
Since $\displaystyle\lim_{h\to 0}\frac{2\sin(h/2)}{h}=1$ and
$\displaystyle\lim_{h\to 0}\cos\big(x+\tfrac{h}{2}\big)=\cos x$, and
$\lim_{h\to 0}\sin(x+h)=\sin x$, we get
$$
f'(x)= -\frac{\cos x}{\sin x\cdot\sin x}
= -\frac{\cos x}{\sin^2 x}.
$$
Write in standard trigonometry form:
$$
f'(x)=-\csc x\cot x.
$$
$$\boxed{,\dfrac{d}{dx}(\csc x) = -\csc x\cot x,}$$
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NCERT Question 11 (v). Find the derivative of $(3 \cot x + 5 \csc x)$
Solution :
Let
$$ f(x) = 3\cot x + 5\csc x. $$
Taking derivative on both sides:
$$ f'(x) = \frac{d}{dx}(3\cot x) + \frac{d}{dx}(5\csc x). $$
Using standard derivatives,
$$ \frac{d}{dx}(\cot x) = -\csc^2 x $$
and
$$ \frac{d}{dx}(\csc x) = -\csc x \cot x. $$
So,
$$
f'(x) = 3(-\csc^2 x) + 5(-\csc x \cot x).
$$
Therefore,
$$
f'(x) = -3\csc^2 x – 5\csc x \cot x.
$$
$$
\boxed{f'(x) = -3\csc^2 x – 5\csc x \cot x}
$$
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NCERT Question 11 (v). Find derivative of $(3\cot x + 5\csc x)$ using first principle
Solution :
Let
$$f(x) = 3 \cot x + 5 \csc x$$
Using first principle,
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
So,
$$f'(x) = \lim_{h \to 0} \frac{3\cot(x+h) + 5\csc(x+h) – 3\cot x – 5\csc x}{h}$$
Separate terms,
$$f'(x) = 3 \lim_{h \to 0} \frac{\cot(x+h) – \cot x}{h} + 5 \lim_{h \to 0} \frac{\csc(x+h) – \csc x}{h}$$
Using known first principle limits,
$$\lim_{h \to 0} \frac{\cot(x+h) – \cot x}{h} = -\csc^2 x$$
$$\lim_{h \to 0} \frac{\csc(x+h) – \csc x}{h} = -\csc x \cot x$$
Substituting,
$$f'(x) = 3(-\csc^2 x) + 5(-\csc x \cot x)$$
Final answer:
$$\boxed{f'(x) = -3\csc^2 x – 5\csc x \cot x}$$
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Question 11(vi). Find the derivative of $5\sin x – 6\cos x + 7$ using first principle
Solution
Let
$$f(x)=5\sin x – 6\cos x + 7.$$
Then
$$f(x+h)=5\sin(x+h)-6\cos(x+h)+7.$$
By the first principle,
$$ f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} $$
$$ f'(x) =\lim_{h\to 0}\frac{5[\sin(x+h)-\sin x]-6[\cos(x+h)-\cos x]}{h}.$$
Use the trigonometric identities
$$\sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right),$$
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).$$
Apply them with $A=x+h,\ B=x$:
$$\sin(x+h)-\sin x=2\cos\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right),$$
$$\cos(x+h)-\cos x=-2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right).$$
Substitute into the limit:
$$
f'(x)=\lim_{h\to 0}\frac{5\cdot 2\cos\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)
-6\cdot\big(-2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)\big)}{h}.
$$
Simplify the numerator:
$$
f'(x)=\lim_{h\to 0}\frac{\big(10\cos\left(x+\tfrac{h}{2}\right)
+12\sin\left(x+\tfrac{h}{2}\right)\big)\sin\left(\tfrac{h}{2}\right)}{h}.
$$
Write $\dfrac{\sin(h/2)}{h}=\dfrac{1}{2}\cdot\dfrac{\sin(h/2)}{h/2}$ to use the standard limit:
$$
f'(x)=\lim_{h\to 0}\left(10\cos\left(x+\tfrac{h}{2}\right)+12\sin\left(x+\tfrac{h}{2}\right)\right)\cdot\frac{1}{2}\cdot\frac{\sin(h/2)}{h/2}.
$$
Now take limits:
$$
\lim_{h\to 0}\frac{\sin(h/2)}{h/2}=1,\qquad
\lim_{h\to 0}\cos\left(x+\tfrac{h}{2}\right)=\cos x,\qquad
\lim_{h\to 0}\sin\left(x+\tfrac{h}{2}\right)=\sin x
$$
Therefore
$$
f'(x)=\frac{1}{2}\big(10\cos x + 12\sin x\big)=5\cos x + 6\sin x.
$$
$$\boxed{f'(x)=5\cos x + 6\sin x}$$
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Question 11(viii) : Find the derivative of $f(x)=2\tan x – 7\sec x$ using first principle
Solution:
Let
$$f'(x)=2\frac{d}{dx}(\tan x)-7\frac{d}{dx}(\sec x).$$
First compute $\dfrac{d}{dx}(\tan x)$ from first principle. Write $\tan x=\dfrac{\sin x}{\cos x}$.
$$
\frac{d}{dx}(\tan x)
=\lim_{h\to 0}\frac{\tan(x+h)-\tan x}{h}
=\lim_{h\to 0}\frac{\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}}{h}.
$$
Combine the fraction:
$$
\frac{\tan(x+h)-\tan x}{h}
=\frac{1}{h}\cdot\frac{\sin(x+h)\cos x-\sin x\cos(x+h)}{\cos(x+h)\cos x}.
$$
Use the identity $\sin A\cos B-\cos A\sin B=\sin(A-B)$ with $A=x+h,\ B=x$:
$$
\sin(x+h)\cos x-\sin x\cos(x+h)=\sin((x+h)-x)=\sin h.
$$
Thus
$$
\frac{\tan(x+h)-\tan x}{h}
=\frac{1}{h}\cdot\frac{\sin h}{\cos(x+h)\cos x}
=\frac{\sin h}{h}\cdot\frac{1}{\cos(x+h)\cos x}.
$$
Taking $h\to0$ and using $\lim_{h\to0}\dfrac{\sin h}{h}=1$ and $\lim_{h\to0}\cos(x+h)=\cos x$,
$$
\frac{d}{dx}(\tan x)=\frac{1}{\cos^2 x}=\sec^2 x.
$$
Next compute $\dfrac{d}{dx}(\sec x)$ from first principle. Write $\sec x=\dfrac{1}{\cos x}$.
$$
\frac{d}{dx}(\sec x)
=\lim_{h\to 0}\frac{\sec(x+h)-\sec x}{h}
=\lim_{h\to 0}\frac{\dfrac{1}{\cos(x+h)}-\dfrac{1}{\cos x}}{h}.
$$
Combine into one fraction:
$$
\frac{\sec(x+h)-\sec x}{h}
=\frac{1}{h}\cdot\frac{\cos x-\cos(x+h)}{\cos(x+h)\cos x}.
$$
Use the identity $\cos A-\cos B=-2\sin!\big(\tfrac{A+B}{2}\big)\sin!\big(\tfrac{A-B}{2}\big)$ with $A=x,\ B=x+h$:
$$
\cos x-\cos(x+h)=2\sin!\left(x+\tfrac{h}{2}\right)\sin!\left(\tfrac{h}{2}\right).
$$
So
$$
\frac{\sec(x+h)-\sec x}{h}
=\frac{2\sin!\left(x+\tfrac{h}{2}\right)}{\cos(x+h)\cos x}\cdot\frac{\sin(h/2)}{h}.
$$
Rewrite $\dfrac{\sin(h/2)}{h}=\dfrac{1}{2}\cdot\dfrac{\sin(h/2)}{(h/2)}$ and take limits. Using $\lim_{u\to0}\dfrac{\sin u}{u}=1$ and $\cos(x+h)\to\cos x$,
$$
\frac{d}{dx}(\sec x)
=\frac{2\sin x}{\cos^2 x}\cdot\frac{1}{2}
=\frac{\sin x}{\cos^2 x}
=\sec x\tan x.
$$
Now combine results:
$$
f'(x)=2\sec^2 x – 7\sec x\tan x.
$$
$$\boxed{,f'(x)=2\sec^2 x – 7\sec x\tan x,}$$
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