Limits and Derivatives

Exercise 12.2 Solutions
Key Derivative Rules:
1. $\frac{d}{dx}(x^n) = nx^{n-1}$
2. First Principle: $f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$
3. Product Rule: $(uv)’ = u’v + uv’$
4. Quotient Rule: $(\frac{u}{v})’ = \frac{u’v – uv’}{v^2}$
Q1-Q3
1. Derivative of $x^2 – 2$ at $x = 10$.
Differentiate $f(x) = x^2 – 2 \Rightarrow f'(x) = 2x$.
At x=10 $f'(10) = 2(10) = 20$.

2. Derivative of $x$ at $x = 1$.
Differentiate $f(x) = x \Rightarrow f'(x) = 1$.
At x=1 $f'(1) = 1$.

3. Derivative of $99x$ at $x = 100$.
Differentiate $f(x) = 99x \Rightarrow f'(x) = 99$.
At x=100 $f'(100) = 99$.
Q4
4. Find derivative from first principle:

(i) $x^3 – 27$

Limit $f'(x) = \lim_{h \to 0} \frac{(x+h)^3 – 27 – (x^3 – 27)}{h}$
Expand $= \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 – x^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2$.

(ii) $(x-1)(x-2) = x^2 – 3x + 2$

Limit $f'(x) = \lim_{h \to 0} \frac{((x+h)^2 – 3(x+h) + 2) – (x^2 – 3x + 2)}{h}$
Simplify $= \lim_{h \to 0} \frac{2xh + h^2 – 3h}{h} = \lim_{h \to 0} (2x + h – 3) = 2x – 3$.

(iii) $\frac{1}{x^2}$

Limit $f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} – \frac{1}{x^2}}{h} = \lim_{h \to 0} \frac{x^2 – (x+h)^2}{h x^2 (x+h)^2}$
Result $= \lim_{h \to 0} \frac{-2xh – h^2}{h x^2 (x+h)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$.

(iv) $\frac{x+1}{x-1}$

Limit $f'(x) = \lim_{h \to 0} \frac{1}{h} [\frac{x+h+1}{x+h-1} – \frac{x+1}{x-1}]$
Simplify Num: $(x+h+1)(x-1) – (x+1)(x+h-1) = -2h$.
Result $\lim_{h \to 0} \frac{-2h}{h(x+h-1)(x-1)} = \frac{-2}{(x-1)^2}$.
Q5-Q8
5. $f(x) = \frac{x^{100}}{100} + \dots + \frac{x^2}{2} + x + 1$. Prove $f'(1) = 100 f'(0)$.
f'(x) $x^{99} + x^{98} + \dots + x + 1$.
f'(1) $1^{99} + \dots + 1 = 100$ terms = 100.
f'(0) $0 + \dots + 0 + 1 = 1$.
$f'(1) = 100 = 100(1) = 100 f'(0)$. Proved.

6. Derivative of $x^n + ax^{n-1} + \dots + a^n$.
Rule Power rule on each term.
$nx^{n-1} + a(n-1)x^{n-2} + \dots + a^{n-1}$

7. Derivatives of:
(i) $(x-a)(x-b) = x^2 – (a+b)x + ab \Rightarrow 2x – (a+b)$.
(ii) $(ax^2+b)^2 = a^2x^4 + 2abx^2 + b^2 \Rightarrow 4a^2x^3 + 4abx$.
(iii) $\frac{x-a}{x-b}$. Quotient rule: $\frac{1(x-b) – 1(x-a)}{(x-b)^2} = \frac{a-b}{(x-b)^2}$.

8. Derivative of $\frac{x^n – a^n}{x – a}$.
Quotient Rule $\frac{nx^{n-1}(x-a) – (x^n-a^n)(1)}{(x-a)^2}$.
$\frac{nx^{n-1}(x-a) – x^n + a^n}{(x-a)^2}$
Q9-Q11
9. Find derivatives:
(i) $2x – \frac{3}{4} \Rightarrow 2$.
(ii) $(5x^3 + 3x – 1)(x – 1)$. Product rule or expand.
Exp: $5x^4 – 5x^3 + 3x^2 – 4x + 1 \Rightarrow 20x^3 – 15x^2 + 6x – 4$.
(iii) $x^{-3}(5+3x) = 5x^{-3} + 3x^{-2} \Rightarrow -15x^{-4} – 6x^{-3}$.
(iv) $x^5(3-6x^{-9}) = 3x^5 – 6x^{-4} \Rightarrow 15x^4 + 24x^{-5}$.
(v) $x^{-4}(3-4x^{-5}) = 3x^{-4} – 4x^{-9} \Rightarrow -12x^{-5} + 36x^{-10}$.
(vi) $\frac{2}{x+1} – \frac{x^2}{3x-1} \Rightarrow \frac{-2}{(x+1)^2} – \frac{2x(3x-1) – x^2(3)}{(3x-1)^2}$.
Simplify 2nd term: $\frac{6x^2-2x-3x^2}{(3x-1)^2} = \frac{3x^2-2x}{(3x-1)^2}$.

10. Derivative of $\cos x$ (First Principle).
Limit $\lim_{h \to 0} \frac{\cos(x+h) – \cos x}{h} = \lim_{h \to 0} \frac{-2\sin(x + h/2)\sin(h/2)}{h}$.
Solve $-2\sin x \cdot \lim_{h \to 0} \frac{\sin(h/2)}{2(h/2)} = -\sin x$.

11. Trigonometric derivatives:
(i) $\sin x \cos x = \frac{1}{2}\sin 2x \Rightarrow \cos 2x$.
(ii) $\sec x \Rightarrow \sec x \tan x$.
(iii) $5\sec x + 4\cos x \Rightarrow 5\sec x \tan x – 4\sin x$.
(iv) $\csc x \Rightarrow -\csc x \cot x$.
(v) $3\cot x + 5\csc x \Rightarrow -3\csc^2 x – 5\csc x \cot x$.
(vi) $5\sin x – 6\cos x + 7 \Rightarrow 5\cos x + 6\sin x$.
(vii) $2\tan x – 7\sec x \Rightarrow 2\sec^2 x – 7\sec x \tan x$.

learncbsehub.in