CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 10

CBSE Class 11 Mathematics Worksheet - Limits and Derivatives. The questions in the worksheets have been specifically designed by best teachers so that the students can practise them to clear their concepts and get better marks in tests and examinations. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Differentiate using first principle method \( f(x) = \cos(3x) \)
Answer: \( \frac{dy}{dx} = \lim_{h \to 0} \left( \frac{f(x + h) - f(x)}{h} \right) \)
\( = \lim_{h \to 0} \left( \frac{\cos(3x + 3h) - \cos(3x)}{h} \right) \)
\( = \lim_{h \to 0} \left( \frac{-2 \sin\left(\frac{6x + 3h}{2}\right) \cdot \sin\left(\frac{3h}{2}\right)}{h} \right) \) {cos A - cos B formula}
\( = \lim_{h \to 0} \left( \frac{-2 \sin\left(\frac{6x + 3h}{2}\right) \cdot \sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \times \frac{3}{2} \right) \)
\( = \lim_{h \to 0} \left( \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \right) \times \left[ -3 \lim_{h \to 0} \left( \sin\left(\frac{6x + 3h}{2}\right) \right) \right] \)
\( = 1 \times (-3 \sin(3x)) \) \( \left\{\lim_{x \to 0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( \frac{dy}{dx} = f'(x) = -3 \sin(3x) \) ans.

Question. Differentiate using first formula \( f(x) = \tan(2x) \)
Answer: \( f'(x) = \frac{dy}{dx} = \lim_{h \to 0} \left( \frac{\tan(2x + 2h) - \tan(2x)}{h} \right) \)
\( = \lim_{h \to 0} \left( \frac{\frac{\sin(2x+2h)}{\cos(2x+2h)} - \frac{\sin(2x)}{\cos(2x)}}{h} \right) \)
\( = \lim_{h \to 0} \left( \frac{\sin(2x+2h)\cos(2x) - \cos(2x+2h)\sin(2x)}{h \cos(2x+2h) \cos(2x)} \right) \)
\( = \lim_{h \to 0} \left( \frac{\sin(2x + 2h - 2x)}{h \cos(2x+2h) \cos(2x)} \right) \) {sin(A - B) formula}
\( = \lim_{h \to 0} \left( \frac{\sin(2h)}{2h \cos(2x+2h) \cos(2x)} \times 2 \right) \)
\( = \lim_{h \to 0} \left( \frac{\sin(2h)}{2h} \right) \times \lim_{h \to 0} \left( \frac{2}{\cos(2x+2h)\cos(2x)} \right) \)
\( = 1 \times \frac{2}{\cos(2x)\cos(2x)} \)
\( \therefore \frac{dy}{dx} = 2 \sec^2(2x) \) ans.

Question. \( f(x) = \sqrt{\tan x} \), find \( f'(x) \) first principle method.
Answer: \( f(x) = \sqrt{\tan x} \)
\( f'(x) = \lim_{h \to 0} \left[ \frac{\sqrt{\tan(x + h)} - \sqrt{\tan x}}{h} \right] \)
Rationalize
\( = \lim_{h \to 0} \left[ \frac{\tan(x+h) - \tan x}{h (\sqrt{\tan(x+h)} + \sqrt{\tan x})} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\tan(x + h - x) \cdot [1 + \tan(x+h)\tan x]}{h (\sqrt{\tan(x+h)} + \sqrt{\tan x})} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\tan ℎ \cdot [1 + \tan(x+h)\tan x]}{h (\sqrt{\tan(x+h)} + \sqrt{\tan x})} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\tan ℎ}{h} \right] \times \lim_{h \to 0} \left[ \frac{1 + \tan(x+h)\tan x}{\sqrt{\tan(x+h)} + \sqrt{\tan x}} \right] \)
\( = 1 \times \frac{1 + \tan^2 x}{2\sqrt{\tan x}} \)
\( \therefore f'(x) = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x \) ans.

Question. \( f(x) = \sec^2 x \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\sec^2(x + h) - \sec^2 x}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\frac{1}{\cos^2(x+h)} - \frac{1}{\cos^2 x}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\cos^2 x - \cos^2(x+h)}{h \cos^2(x+h) \cos^2 x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\{\cos x + \cos(x+h)\}\{\cos x - \cos(x+h)\}}{h \cos^2(x+h) \cos^2 x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\{\cos x + \cos(x+h)\}\left\{-2 \sin\left(\frac{2x+h}{2}\right) \cdot \sin\left(-\frac{h}{2}\right)\right\}}{h \cos^2(x+h) \cos^2 x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2 \sin\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right) \times \{\cos x + \cos(x+h)\}}{2 \times \frac{h}{2} \cdot \cos^2(x+h) \cos^2 x} \right] \)
\( = \lim_{h \to 0} \left( \sin\left(\frac{2x+h}{2}\right) \right) \times \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \times \lim_{h \to 0} \left[ \frac{\cos x + \cos(x+h)}{\cos^2(x+h) \cos^2 x} \right] \)
\( = \sin(x) \times 1 \times \frac{\cos x + \cos x}{\cos^2 x \cdot \cos^2 x} \)
\( = \frac{\sin x \cdot 2\cos x}{\cos^2 x \cdot \cos^2 x} \)
\( = \frac{2 \sin x}{\cos x} \cdot \frac{1}{\cos^2 x} \)
\( f'(x) = 2 \tan x \cdot \sec^2 x \) ans.

Question. \( f(x) = \sin(x^2) \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\sin(x + h)^2 - \sin(x^2)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(x^2 + h^2 + 2hx) - \sin(x^2)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2 \cos\left(\frac{2x^2 + h^2 + 2hx}{2}\right) \cdot \sin\left(\frac{h^2 + 2hx}{2}\right)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2 \cos\left(\frac{2x^2 + h^2 + 2hx}{2}\right) \cdot \sin\left(\frac{h^2 + 2hx}{2}\right)}{h \cdot \left(\frac{h^2+2hx}{2}\right)} \times \left(\frac{h^2+2hx}{2}\right) \right] \)
\( = \lim_{h \to 0} \left( \frac{\sin\left(\frac{h^2 + 2hx}{2}\right)}{\frac{h^2 + 2hx}{2}} \right) \times \lim_{h \to 0} \left[ 2 \cos\left(\frac{2x^2 + h^2 + 2hx}{2}\right) \right] \times \lim_{h \to 0} \left( \frac{h^2 + 2hx}{2h} \right) \)
\( = 1 \times (2\cos(x^2)) \times \lim_{h \to 0} \left[ \frac{h(h+2x)}{2h} \right] \)
\( = 2 \cos(x^2) \times \left( \frac{2x}{2} \right) \)
\( \therefore \frac{dy}{dx} = 2x \cdot \cos(x^2) \) ans.

Question. \( f(x) = \tan\sqrt{x} \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\tan(\sqrt{x + h}) - \tan\sqrt{x}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\frac{\sin(\sqrt{x+h})}{\cos(\sqrt{x+h})} - \frac{\sin\sqrt{x}}{\cos\sqrt{x}}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(\sqrt{x+h})\cos\sqrt{x} - \cos(\sqrt{x+h})\sin\sqrt{x}}{h \cos(\sqrt{x+h})\cos\sqrt{x}} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(\sqrt{x+h} - \sqrt{x})}{h \cos(\sqrt{x+h})\cos\sqrt{x}} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(\sqrt{x+h} - \sqrt{x}) \times (\sqrt{x+h} - \sqrt{x})}{h (\sqrt{x+h} - \sqrt{x}) \cos(\sqrt{x+h}) \cos\sqrt{x}} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(\sqrt{x+h} - \sqrt{x})}{\sqrt{x+h} - \sqrt{x}} \right] \times \lim_{h \to 0} \left[ \frac{1}{\cos(\sqrt{x+h})\cos\sqrt{x}} \right] \times \lim_{h \to 0} \left[ \frac{\sqrt{x+h} - \sqrt{x}}{h} \right] \)
\( = 1 \times \frac{1}{\cos\sqrt{x}\cos\sqrt{x}} \times \lim_{h \to 0} \left[ \frac{x+h-x}{h(\sqrt{x+h} + \sqrt{x})} \right] \)
\( = \sec^2\sqrt{x} \times \frac{1}{\sqrt{x} + \sqrt{x}} \)
\( \therefore \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cdot \sec^2\sqrt{x} \) ans.

Question. \( f(x) = x \cos x \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{(x + h) \cdot \cos(x + h) - x \cos x}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x \cos(x+h) + h \cos(x+h) - x \cos x}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x\{\cos(x+h) - \cos x\} + h \cos(x+h)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x \left\{ -2 \sin\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right) \right\} + h \cos(x+h)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{-2x \cdot \sin\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right)}{2 \times \frac{h}{2}} + \frac{h \cos(x+h)}{h} \right] \)
\( = \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \times \left[ x \lim_{h \to 0} \left( -\sin\left(\frac{2x+h}{2}\right) \right) \right] + \lim_{h \to 0} (\cos(x + h)) \)
\( = (1)(-x \sin x) + \cos x \)
\( \therefore f'(x) = -x \sin x + \cos x \) ans.

Question. \( f(x) = \frac{\sin x}{x} \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\frac{\sin(x+h)}{x+h} - \frac{\sin x}{x}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x \sin(x+h) - (x+h)\sin x}{h(x+h)x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x \sin(x+h) - x \sin x - h \sin x}{h(x+h)x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x\{\sin(x+h) - \sin x\} - h \sin x}{h(x+h)x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x\{\sin(x+h) - \sin x\}}{h(x+h)x} - \frac{h \sin x}{h(x+h)x} \right] \)
\( = \lim_{h \to 0} \left[ \frac{x \cdot 2 \cos\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right)}{2 \times \frac{h}{2}(x+h)x} - \frac{\sin x}{(x+h)x} \right] \)
\( = \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \times \lim_{h \to 0} \left( \frac{x \cos\left(\frac{2x+h}{2}\right)}{(x+h)x} \right) - \lim_{h \to 0} \left( \frac{\sin x}{(x+h)x} \right) \)
\( = 1 \times \left( \frac{x \cos x}{x^2} \right) - \frac{\sin x}{x^2} \)
\( \therefore \frac{dy}{dx} = \frac{x \cos x - \sin x}{x^2} \) ans.

Question. \( f(x) = \sin x - \cos x \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\{\sin(x + h) - \cos(x + h)\} - \{\sin x - \cos x\}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\{\sin(x+h) - \sin x\} - \{\cos(x+h) - \cos x\}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2 \cos\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right) + 2 \sin\left(\frac{2x+h}{2}\right) \cdot \sin\left(\frac{h}{2}\right)}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2 \sin\left(\frac{h}{2}\right)\left\{\cos\left(\frac{2x+h}{2}\right) + \sin\left(\frac{2x+h}{2}\right)\right\}}{2 \times \frac{h}{2}} \right] \)
\( = \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \times \lim_{h \to 0} \left( \cos\left(\frac{2x+h}{2}\right) + \sin\left(\frac{2x+h}{2}\right) \right) \)
\( = 1 \times (\cos x + \sin x) \)
\( \therefore f'(x) = \cos x + \sin x \) ans.

Question. \( f(x) = \frac{2x^2 + 1}{x - 3} \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\left\{ \frac{2(x + h)^2 + 1}{x + h - 3} \right\} - \left\{ \frac{2x^2 + 1}{x - 3} \right\}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\left( \frac{2x^2 + 2h^2 + 4hx + 1}{x+h-3} \right) - \frac{2x^2 + 1}{x - 3}}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{(2x^2 + 2h^2 + 4hx + 1)(x - 3) - (2x^2 + 1)(x + h - 3)}{h(x+h-3)(x-3)} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2x^3 - 6x^2 + 2h^2x - 6h^2 + 2hx^2 - 6hx + x - 3 - 2x^3 - 2x^2h + 6x^2 + x - h + 3}{h(x+h-3)(x-3)} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2h^2x - 6h^2 - 6hx - h}{h(x+h-3)(x-3)} \right] \)
\( = \lim_{h \to 0} \left[ \frac{h(2hx - 6h - 6x - 1)}{h(x+h-3)(x-3)} \right] \)
\( \therefore f'(x) = \frac{-6x - 1}{(x-3)^2} \) ans.