Derivatives: Part 10: IITJEE maths tutorial problems for practice

Problem 1: If x=3\cos{\theta}-\cos^{3}{\theta}, and y=3\sin{\theta}-\sin^{3}{\theta}, then \frac{dy}{dx} is equal to:

(a) -\cot^{3}{\theta} (b) -\tan^{3}{\theta} (c) \cot^{3}{\theta} (d) \tan^{3}{\theta}

Problem 2: If x = \tan{\theta} + \cot{\theta}, and y=2 \log{(\cot{\theta})}, then \frac{dy}{dx} is equal to:

(a) \tan{(2\theta)} (b) \cot{(2\theta)} (c) \tan{\theta} (d) \sec^{2}{2\theta}

Problem 3: \frac{d}{dx}\log{\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}} is equal to:

(a) \tan{\frac{x}{2}} (b) \sin{x} (c) cosec(x) (d) \tan{x}

Problem 4: y=2^{2(\log_{2}{(x+2)}-\log_{2}{(x+1)})}, then \frac{dy}{dx} is:

(a) \frac{-2(x+2)}{(x+1)^{3}} (b) \frac{4(x+2)}{(x+1)^{3}} (c) \frac{2(x+2)}{(x+1)^{3}} (d) \frac{-6(x+2)}{(x+1)^{3}}

Problem 5: \frac{d}{dx}(\arctan{\sqrt{\frac{e^{x}-1}{e^{x}+1}}}) is equal to:

(a) \frac{1}{1+e^{2x}} (b) \frac{1}{2\sqrt{e^{2x}-1}} (c) \frac{e^{x}}{2\sqrt{1+e^{2x}}} (d) \frac{1}{2\sqrt{1-e^{2x}}}

Problem 6: y=e^{m\arcsin{x}} then (1-x^{2}) (y^{'})^{2} is equal to :

(a) y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}y^{2}

Problem 7: If y = \sin(m \arcsin{x}) then (1-x^{2})(\frac{dy}{dx})^{2} is

(a) m^{2}y^{2} (b) m^{2}(1-y^{2}) (c) -m^{2}y^{2} (d) m^{2}(1+y^{2})

Problem 8: \frac{d}{dx}(\cos{\arctan{x}}) is:

(a) \frac{1}{2\sqrt{1+x^{2}}} (b) \frac{-x}{(1+x^{2})^{\frac{3}{2}}}

(c) \frac{-x}{\sqrt{1+x^{2}}} (d) \frac{2x}{\sqrt{1+x^{2}}}

Problem 9: If y = \arcsin{\frac{a\cos{x}+b\sin{x}}{\sqrt{a^{2}+b^{2}}}} then \frac{d^{2}y}{dx^{2}} is:

(a) -1 (b) 0 (c) 1 (d) \arctan{\frac{b}{a}}

Problem 10: \arctan{\frac{4x}{4-x^{2}}} is:

(a) \frac{1}{4-x^{2}} (b) \frac{1}{4+x^{2}} (c) \frac{4}{4+x^{2}} (d) \frac{4}{4-x^{2}}

Regards,

Nalin Pithwa.

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