Exercise 5.2, Question 1
Differentiate \(\sin(x^2 + 5)\).
Show solution
- Outer sin, inner \(x^2 + 5\): \(\cos(x^2 + 5) \cdot 2x\).
Answer: \(2x\cos(x^2 + 5)\)
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2
Exercise 5.2: Chain rule; where derivatives fail. Chain rule: \(\dfrac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)\), working from the outside in. A function is differentiable at c only if the left and right derivatives agree there; a corner (like \(|x - 1|\) at 1) or a jump (like \([x]\) at an integer) spoils this.
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Exercise 5.2 questions and solutions
Exercise 5.2, Question 2
Differentiate \(\cos(\sin x)\).
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- \(-\sin(\sin x) \cdot \cos x\).
Answer: \(-\cos x\,\sin(\sin x)\)
Exercise 5.2, Question 3
Differentiate \(\sin(ax + b)\).
Show solution
- \(\cos(ax + b) \cdot a\).
Answer: \(a\cos(ax + b)\)
Exercise 5.2, Question 4
Differentiate \(\sec(\tan\sqrt x)\).
Show solution
- Three layers: \(\sec u \to \sec u\tan u\), \(u = \tan v \to \sec^2 v\), \(v = \sqrt x \to \dfrac{1}{2\sqrt x}\).
Answer: \[\dfrac{\sec(\tan\sqrt x)\tan(\tan\sqrt x)\sec^2\sqrt x}{2\sqrt x}\]
Exercise 5.2, Question 5
Differentiate \(\dfrac{\sin(ax + b)}{\cos(cx + d)}\).
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- Quotient rule: \[\dfrac{a\cos(ax + b)\cos(cx + d) - \sin(ax + b)\cdot(-c\sin(cx + d))}{\cos^2(cx + d)}\]
Answer: \[\dfrac{a\cos(ax + b)\cos(cx + d) + c\sin(ax + b)\sin(cx + d)}{\cos^2(cx + d)}\]
Exercise 5.2, Question 6
Differentiate \(\cos x^3 \cdot \sin^2(x^5)\).
Show solution
- Product rule: \((\cos x^3)' = -3x^2\sin x^3\); \[(\sin^2 x^5)' = 2\sin x^5\cos x^5 \cdot 5x^4\]
Answer: \[10x^4\sin x^5\cos x^5\cos x^3 - 3x^2\sin x^3\sin^2 x^5\]
Exercise 5.2, Question 7
Differentiate \(2\sqrt{\cot(x^2)}\).
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- \[2 \cdot \dfrac{1}{2\sqrt{\cot x^2}} \cdot (-\csc^2 x^2) \cdot 2x\]
Answer: \(-\dfrac{2x\csc^2(x^2)}{\sqrt{\cot(x^2)}}\)
Exercise 5.2, Question 8
Differentiate \(\cos\sqrt x\).
Show solution
- \(-\sin\sqrt x \cdot \dfrac{1}{2\sqrt x}\).
Answer: \(-\dfrac{\sin\sqrt x}{2\sqrt x}\)
Exercise 5.2, Question 9
Prove \(f(x) = |x - 1|\) is not differentiable at \(x = 1\).
Show solution
- Left derivative: \[\begin{aligned}\lim_{h \to 0^-}\dfrac{|h| - 0}{h} &= \lim_{h \to 0^-}\dfrac{-h}{h} \\ &= -1\end{aligned}\]
- Right derivative: \(\lim_{h \to 0^+}\dfrac{|h|}{h} = 1\).
- They differ, so \(f'(1)\) does not exist.
Answer: Proved
Exercise 5.2, Question 10
Prove \(f(x) = [x]\), \(0 < x < 3\), is not differentiable at \(x = 1\) and \(x = 2\).
Show solution
- At 1: for small \(h < 0\), \[\begin{aligned}\dfrac{[1 + h] - [1]}{h} &= \dfrac{0 - 1}{h} \\ &= -\dfrac1h \to +\infty\end{aligned}\], so the left derivative does not exist. (The right derivative is \(\dfrac{1 - 1}{h} = 0\).)
- At 2 the same happens: \(\dfrac{[2 + h] - 2}{h} = -\dfrac1h\) for small \(h < 0\).
- (Also, f is not even continuous at 1 and 2, and a differentiable function must be continuous.)
Answer: Proved
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