CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 11

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. \( f(x) = \sqrt{2x + 3} \) Using first principle method.
Answer: \( f'(x) = \lim_{h \to 0} \left[ \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h} \right] \)
Rationalize
\( = \lim_{h \to 0} \left[ \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h} \times \frac{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}}{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}} \right] \)
\( = \lim_{h \to 0} \left[ \frac{(2x + 2h + 3) - (2x + 3)}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2h}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2}{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}} \right] \)
\( = \lim_{h \to 0} \left[ \frac{2}{2\sqrt{2x + 3}} \right] \)
\( \therefore \frac{dy}{dx} = \frac{1}{\sqrt{2x + 3}} \) ans.

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

Question. Differentiate w. r. t x (product rule)
\( f(x) = (ax + b)^m(cx + d)^n \)
Answer: We have \( f(x) = (ax + b)^m(cx + d)^n \)
Differentiate both sides w.r.t. \( x \) (product rule)
\( f'(x) = (ax + b)^m \cdot \frac{d}{dx}(cx + d)^n + (cx + d)^n \cdot \frac{d}{dx}(ax + b)^m \)
\( = (ax + b)^m \cdot n(cx + d)^{n-1} \cdot \frac{d}{dx}(cx + d) + (cx + d)^n \cdot m(ax + b)^{m-1} \cdot \frac{d}{dx}(ax + b) \)
\( = (ax + b)^m \cdot n(cx + d)^{n-1} \cdot (c) + (cx + d)^n \cdot m(ax + b)^{m-1} \cdot (a) \)
\( = (ax + b)^{m-1} \cdot (cx + d)^{n-1} [(ax + b)nc + (cx + d)ma] \)
\( = (ax + b)^{m-1} \cdot (cx + d)^{n-1} [ancx + bnc + cmax + dma] \)
\( f'(x) = (ax + b)^{m-1} \cdot (cx + d)^{n-1} [ax(nc + mc) + (bnc + dma)] \) ans.

Question. Differentiate w. r. t x (product rule)
\( f(x) = (x + \sec x)(x - \tan x) \)
Answer: We have \( f(x) = (x + \sec x)(x - \tan x) \)
Differentiate both sides w.r.t. \( x \) (product rule)
\( f'(x) = (x + \sec x) \cdot \frac{d}{dx}(x - \tan x) + (x - \tan x)\frac{d}{dx}(x + \sec x) \)
\( = (x + \sec x)(1 - \sec^2 x) + (x - \tan x)(1 + \sec x \tan x) \) ans.

Question. Differentiate w. r. t x (product rule)
\( f(x) = (x \sin x + \cos x)(x \cos x - \sin x) \)
Answer: We have \( f(x) = (x \sin x + \cos x)(x \cos x - \sin x) \)
Differentiate both sides w.r.t. \( x \) (product rule)
\( f'(x) = (x \sin x + \cos x) \cdot \frac{d}{dx}(x \cos x - \sin x) + (x \cos x - \sin x) \cdot \frac{d}{dx}(x \sin x + \cos x) \)
\( = (x \sin x + \cos x) \left[ x \cdot \frac{d}{dx}(\cos x) + \cos x \cdot \frac{d}{dx}(x) - \frac{d}{dx}(\sin x) \right] + (x \cos x - \sin x) \left[ x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) \right] \)
\( = (x \sin x + \cos x)(-x \sin x + \cos x - \cos x) + (x \cos x - \sin x)(x \cos x + \sin x - \sin x) \)
\( = (x \sin x + \cos x)(-x \sin x) + (x \cos x - \sin x)(x \cos x) \)
\( = -x^2 \sin^2 x - x \sin x \cos x + x^2 \cos^2 x - x \sin x \cos x \)
\( = x^2(\cos^2 x - \sin^2 x) - 2x \sin x \cos x \)
\( f'(x) = x^2 \cos(2x) - x \sin(2x) \) ans.

Question. Differentiate w. r. t x (product rule)
\( f(x) = x^{-4}(3 - 4x^{-5}) \)
Answer: Differentiate w.r.t. \( x \) (product rule)
\( f'(x) = x^{-4} \cdot \frac{d}{dx}(3 - 4x^{-5}) + (3 - 4x^{-5}) \cdot \frac{d}{dx}(x^{-4}) \)
\( = x^{-4}(0 + 20x^{-6}) + (3 - 4x^{-5})(-4x^{-5}) \)
\( = x^{-4}(20x^{-6}) - (3 - 4x^{-5})(4x^{-5}) \)
\( = 20x^{-10} - 12x^{-5} + 16x^{-10} \)
\( = 36x^{-10} - 12x^{-5} \) ans.

Question. Differentiate w. r. t x (product rule)
\( f(x) = \frac{1 - \tan x}{1 + \tan x} \)
Answer: We have \( f(x) = \frac{1-\tan x}{1+\tan x} \)
\( f'(x) = \frac{1 - \tan x}{1 + \tan x} \)
\( = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}} \)
\( = \frac{\cos x - \sin x}{\cos x + \sin x} \)
Differentiate w.r.t. \( x \) (quotient rule)
\( f'(x) = \frac{(\cos x + \sin x)\frac{d}{dx}(\cos x - \sin x) - (\cos x - \sin x)\frac{d}{dx}(\cos x + \sin x)}{(\cos x + \sin x)^2} \)
\( = \frac{(\cos x + \sin x)(-\sin x - \cos x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2} \)
\( = \frac{-(\cos x + \sin x)(\cos x + \sin x) - (\cos x - \sin x)(\cos x - \sin x)}{(\cos x + \sin x)^2} \)
\( = \frac{-(\cos^2 x + \sin^2 x + 2\sin x \cos x) - (\cos^2 x + \sin^2 x - 2\sin x \cos x)}{(\cos x + \sin x)^2} \)
\( = \frac{-(1 + \sin(2x)) - (1 - \sin(2x))}{(\cos x + \sin x)^2} \)
\( f'(x) = \frac{-2}{(\cos x + \sin x)^2} \) ans.

Question. Differentiate w. r. t x (product rule)
\( f(x) = \frac{x}{\sin^n x} \)
Answer: We have \( f(x) = \frac{x}{\sin^n x} \)
Differentiate w.r.t. \( x \) (quotient rule)
\( \frac{dy}{dx} = \frac{\sin^n x \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin^n x)}{(\sin^n x)^2} \)
\( = \frac{\sin^n x \cdot (1) - x \cdot n \cdot \sin^{n-1} x \cdot \frac{d}{dx}(\sin x)}{\sin^{2n} x} \)
\( = \frac{\sin^n x - nx \cdot \sin^{n-1} x \cdot \cos x}{\sin^{2n} x} \)
\( = \frac{\sin^{n-1} x (\sin x - nx \cos x)}{\sin^{2n} x} \)
\( = \frac{\sin x - nx \cos x}{\sin^{2n - n + 1} x} \)
\( f'(x) = \frac{\sin x - nx \cos x}{\sin^{n+1} x} \) ans.

Question. \( f(x) = \frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x} \)
Answer: We have \( \frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x} \)
\( = f(x) = \frac{1}{\sqrt{2}} \cdot \frac{x^2}{\sin x} \)
Differentiate w.r.t. \( x \) (quotient rule)
\( f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{\sin x \cdot \frac{d}{dx}(x^2) - (x^2)\frac{d}{dx}(\sin x)}{(\sin x)^2} \right] \)
\( = \frac{1}{\sqrt{2}} \left[ \frac{\sin x \cdot (2x) - x^2 \cos x}{\sin^2 x} \right] \)
\( f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{2x \sin x - x^2 \cos x}{\sin^2 x} \right] \) ans.

Question. \( f(x) = \frac{\sin x - x \cos x}{x \sin x + \cos x} \)
Answer: We have \( \frac{\sin x - x \cos x}{x \sin x + \cos x} \)
Differentiate w.r.t. \( x \) (quotient rule)
\( f'(x) = \frac{(x \sin x + \cos x) \cdot \frac{d}{dx}(\sin x - x \cos x) - (\sin x - x \cos x)\frac{d}{dx}(x \sin x + \cos x)}{(x \sin x + \cos x)^2} \)
\( = \frac{(x \sin x + \cos x)\left[ \frac{d}{dx}(\sin x) - \left(x \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(x)\right) \right] - (\sin x - x \cos x)\left[ x \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) \right]}{(x \sin x + \cos x)^2} \)
\( = \frac{(x \sin x + \cos x)[\cos x - (-x \sin x + \cos x)] - (\sin x - x \cos x)(x \cos x + \sin x - \sin x)}{(x \sin x + \cos x)^2} \)
\( = \frac{(x \sin x + \cos x)(x \sin x) - (\sin x - x \cos x)(x \cos x)}{(x \sin x + \cos x)^2} \)
\( = \frac{x^2 \sin^2 x + x \sin x \cos x - x \sin x \cos x + x^2 \cos^2 x}{(x \sin x + \cos x)^2} \)
\( = \frac{x^2(\sin^2 x + \cos^2 x)}{(x \sin x + \cos x)^2} \)
\( f'(x) = \frac{x^2}{(x \sin x + \cos x)^2} \) ans.