Derivatives: Part 9: IITJEE maths tutorial problems practice

Problem 1: $\frac{d}{dx}((\frac{1}{b}\arctan{\frac{x}{b}})-\frac{1}{a}\arctan{(\frac{x}{a})})$ is equal to:

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

Problem 2: $\frac{d}{dx}(\frac{x}{2} + \frac{1}{2}\log{(\sin{x}+\cos{x})})$ is equal to:

(a) $\frac{\tan{x}}{1+\tan{x}}$ (b) $\frac{1}{1+\cot{x}}$ (c) $\frac{1-\tan{x}}{1+\tan{x}}$ (d) $\frac{1}{1+\tan{x}}$

Problem 3: If $y=\sqrt{\frac{cosec{x}-\cot{x}}{cosec{x}+\cot{x}}}$ where $0<x<\frac{\pi}{2}$ , then $\frac{dy}{dx}$ is given by :

(a) $cosec{x}(cosec{x}-\cot{x})$ (b) $cosec{x}(\cot{x}-cosec{x})$ (c) $cosec{x}(\cot{x}-cosec{x})$ (d) $\cot{x}(cosec{x}-\cot{x})$

Problem 4: $\frac{d}{dx}\log {|\sec{(x-\frac{\pi}{4})}+\tan{(x-\frac{\pi}{4})}|}$ is equal to:

(a) $\frac{\sqrt{2}}{\sin{x}-\cos{x}}$ (b) $\frac{\sin{x}}{\sin{x}+\cos{x}}$ (c) $\frac{\sqrt{2}}{\sin{x}+\cos{x}}$ (d) $\frac{1}{\sin{x}+\cos{x}}$

Problem 5:

If $r=a(1+\cos{\theta})$ , and $\tan{\phi}=r\frac{d\theta}{dr}$ , then $\phi$ is equal to:

(a) $\frac{-2}{\theta}$ (b) $\frac{\pi}{2} + \frac{\theta}{2}$ (c) $-\frac{\theta}{2}$ (d) $\frac{\pi}{2} - \frac{\theta}{2}$

Problem 6: $\frac{d}{dx}\log{(\sqrt{x+ \sqrt{x^{2}+a^{2}}})}$ is equal to:

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

Problem 7: $\frac{d}{dx}(\log{(1+\sin{(2x)})} + 2 \log{\sec{(\frac{\pi}{4}-x)}})$ is equal to

(a) 0 (b) $\log{2}$ (c) $\frac{4(\cos{x}-\tan{x})}{\sin{x}+\cos{x}}$ (d) $\frac{2\cos{(2x)}}{1+\sin{(2x)}} + \tan{(\frac{\pi}{4}-x)}$

Problem 8: If $x^{2}+xy+y^{2}=1$ , then $\frac{dy}{dx}$ is equal to:

(a) $-\frac{x+2y}{y+2x}$ (b) $-\frac{y+2x}{x+2y}$ (c) $\frac{y+2x}{x+2y}$ (d) $\frac{2(x+y)}{y-2x}$

Problem 9: $\frac{d}{dx}(\arcsin{(\sqrt{\frac{1-x}{2}})})$ is equal to:

(a) $\frac{1}{\sqrt{1-x^{2}}}$ (b) $\frac{-1}{\sqrt{1-x^{2}}}$ (c) $\frac{1}{2\sqrt{1-x^{2}}}$ 9d) $\frac{-1}{2\sqrt{1-x^{2}}}$

Problem 10: If $y = \arctan{(\frac{3a^{2}x-x^{3}}{x^{3}-3ax^{2}})}$ then $\frac{dy}{dx}$ is equal to:

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

Cheers,

Nalin Pithwa

This entry was written by Nalin Pithwa, posted on January 17, 2021 at 11:24 pm, filed under calculus, IITJEE Mains. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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