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})}

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

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