## 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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