Differentiation (SAQs)
Maths-1B | 9. Differentiation – SAQs:
Welcome to SAQs in Chapter 9: Differentiation. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
SAQ-1 : Find the derivative of sin2x from the first principle
$$f(x) = \sin 2x$$
$$f(x + h) = \sin [2(x + h)] = \sin (2x + 2h)$$
$$f(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f(x) = Lt_{h \to 0} \frac{\sin(2x + 2h) – \sin 2x}{h}$$
$$f(x) = Lt_{h \to 0} \frac{1}{h}(2 \cos\left(\frac{(2x + 2h) + 2x}{2}\right) \sin\left(\frac{(2x + 2h) – 2x}{2}\right))$$
$$f(x) = 2 Lt_{h \to 0} \frac{1}{h} \cos \left(2x + h\right) \sin(h)$$
$$f(x) = 2 Lt_{h \to 0} \cos(2x + h) \cdot Lt_{h \to 0} \frac{\sin h}{h}$$
$$f(x) = 2 \cos(2x + 0)(1) = 2 \cos 2x$$
SAQ-2 : Find the derivative of cosax from the first principle
$$f(x) = \cos ax$$
$$f(x + h) = \cos[a(x + h)] = \cos(ax + ah)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\cos(ax + ah) – \cos(ax)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h}[-2 \sin\left(\frac{(ax + ah) + ax}{2}\right) \sin\left(\frac{(ax + ah) – ax}{2}\right)]$$
$$f'(x) = -2 Lt_{h \to 0} \frac{1}{h}[\sin(2ax + \frac{ah}{2}) \sin(\frac{ah}{2})]$$
$$f'(x) = -2 Lt_{h \to 0} \sin(ax + \frac{ah}{2}) \cdot Lt_{h \to 0} \frac{\sin(\frac{ah}{2})}{h}$$
$$f'(x) = -2 \sin(ax + 0) \cdot \left(\frac{a}{2}\right)$$
$$f'(x) = -a \sin ax$$
SAQ-3 : Find the derivative of tan2x from the first principle
$$f(x) = \tan(2x)$$
$$f(x + h) = \tan[2(x + h)] = \tan(2x + 2h)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\tan(2x + 2h) – \tan(2x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left[\frac{\sin(2x + 2h)\cos(2x) – \cos(2x + 2h)\sin(2x)}{\cos(2x + 2h)\cos(2x)}\right]$$
$$f'(x) = Lt_{h \to 0} \frac{\sin 2h}{h} \cdot Lt_{h \to 0} \frac{1}{\cos(2x + 2h)\cos(2x)}$$
$$f'(x) = 2 \cdot \frac{1}{\cos^2(2x)} = 2 \sec^2(2x)$$
SAQ-4 : Find the derivative of cotx from the first principle
$$f(x) = \cot x$$
$$f(x + h) = \cot(x + h)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\cot(x + h) – \cot(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left[\frac{\cos(x + h)\sin x – \sin(x + h)\cos x}{\sin(x + h)\sin x}\right]$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left[-\frac{\sin((x + h) – x)}{\sin(x + h)\sin x}\right]$$
$$f'(x) = Lt_{h \to 0} \frac{-\sin h}{h} \cdot Lt_{h \to 0} \frac{1}{\sin(x + h)\sin x}$$
$$f'(x) = -1 \cdot \frac{1}{\sin^2 x} = -\csc^2 x$$
SAQ-5 : Find the derivative of cos2x from the first principle
$$f(x) = \cos^2 x$$
$$f(x + h) = \cos^2(x + h)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\cos^2(x + h) – \cos^2 x}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{-\sin[(x + h) + x] \sin[(x + h) – x]}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{-\sin(2x + h) \sin(h)}{h}$$
$$f'(x) = -Lt_{h \to 0} \sin(2x + h) \cdot Lt_{h \to 0} \frac{\sin h}{h}$$
$$f'(x) = -\sin(2x) \cdot 1 = -\sin 2x$$
SAQ-6 : Find the derivative of sec3x using first principle
$$f(x) = \sec 3x$$
$$f(x + h) = \sec[3(x + h)] = \sec(3x + 3h)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\sec(3x + 3h) – \sec 3x}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left( \frac{1}{\cos(3x + 3h)} – \frac{1}{\cos 3x} \right)$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left( \frac{\cos 3x – \cos(3x + 3h)}{\cos(3x + 3h) \cos 3x} \right)$$
$$f'(x) = Lt_{h \to 0} \frac{1}{h} \left( \frac{2\sin\left(\frac{6x + 3h}{2}\right) \sin\left(\frac{3h}{2}\right)}{\cos(3x + 3h) \cos 3x} \right)$$
$$f'(x) = 2 Lt_{h \to 0} \frac{\sin(3x + \frac{3h}{2}) \sin(\frac{3h}{2})}{h \cos(3x + 3h) \cos 3x}$$
$$f'(x) = 2 \cdot \frac{\sin 3x}{\cos 3x \cos 3x} \cdot 3 \cdot \frac{1}{2} = 3 \frac{\sin 3x}{\cos^2 3x} = 3 \sec 3x \tan 3x$$
SAQ-7 : Find the derivative of xsinx from the first principle
$$f(x) = x \sin x$$
$$f(x + h) = (x + h) \sin(x + h) = x \sin(x + h) + h \sin(x + h)$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{x \sin(x + h) + h \sin(x + h) – x \sin x}{h}$$
$$f'(x) = Lt_{h \to 0} \left[ x \frac{\sin(x + h) – \sin x}{h} + \frac{h \sin(x + h)}{h} \right]$$
$$f'(x) = Lt_{h \to 0} \left[ x \frac{2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} + \sin(x + h) \right]$$
$$f'(x) = Lt_{h \to 0} \left[ x \cdot 2 \cos(x + \frac{h}{2}) \frac{\sin(\frac{h}{2})}{h} + \sin(x + h) \right]$$
$$f'(x) = 2x \cos(x + 0) \cdot 1 + \sin x = x \cos x + \sin x$$
SAQ-8 : Find the derivative of x3 from the first principle
$$f(x) = x^3$$
$$f(x + h) = (x + h)^3$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f(x + h) = x^3 + 3x^2h + 3xh^2 + h^3$$
$$f'(x) = Lt_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 – x^3}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$
$$f'(x) = Lt_{h \to 0} (3x^2 + 3xh + h^2)$$
$$f'(x) = 3x^2 + 3x \cdot 0 + 0^2 = 3x^2$$
SAQ-9 : Find the derivative of √x + 1 from the first principle
$$f(x) = \sqrt{x + 1}$$
$$f(x + h) = \sqrt{x + h + 1}$$
$$f'(x) = Lt_{h \to 0} \frac{f(x + h) – f(x)}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\sqrt{x + h + 1} – \sqrt{x + 1}}{h}$$
$$f'(x) = Lt_{h \to 0} \frac{\sqrt{x + h + 1} – \sqrt{x + 1}}{h} \times \frac{\sqrt{x + h + 1} + \sqrt{x + 1}}{\sqrt{x + h + 1} + \sqrt{x + 1}}$$
$$f'(x) = Lt_{h \to 0} \frac{(x + h + 1) – (x + 1)}{h(\sqrt{x + h + 1} + \sqrt{x + 1})}$$
$$f'(x) = Lt_{h \to 0} \frac{h}{h(\sqrt{x + h + 1} + \sqrt{x + 1})}$$
$$f'(x) = Lt_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}}$$
$$f'(x) = \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} = \frac{1}{2\sqrt{x + 1}}$$