Derivatives: part 8: IITJEE mains tutorial problems practice

Problem 1: If y=b(\arctan{(\frac{x}{y})})+ \arctan{(\frac{y}{x})}, then \frac{dy}{dx} is equal to:

(a) \frac{x}{2} (b) -1 (c) 0 (d) b

Problem 2: If r=a(1+\cos{\theta}), then \sqrt{r^{2}+(\frac{dr}{d\theta})^{2}} is:

(a) 2a\cos{\theta} (b) 2a \sin{(\frac{\theta}{2})} (c) 2a \cos{(\frac{\theta}{2})} (d) 2a \sin{\theta}

Problem 3: \frac{d}{dx}\arctan{\log_{10}{x}} is equal to:

(a) \frac{1}{1 + (\log_{10}{x})^{2}} (b) \frac{1}{x \log_{10}{(1+ (\log_{10}{x})^{2})}} (c) \frac{1}{x(1+(\log_{10}{x})^{2})} (d) \frac{1}{10 \log{x}(1+(\log_{10}{x})^{2})}

Problem 4: If \sin^{2}(mx) + \cos^{2}(ny)=a^{2}, then \frac{dy}{dx} is equal to:

(a) \frac{m \sin{(2mx)}}{n \sin{(2ny)}} (b) \frac{n\sin{(2mx)}}{m\sin{(2ny)}} (c) \frac{n\sin{(2ny)}}{m\sin{(2mx)}} (d) \frac{-m\sin{(2mx)}}{n\sin{(2ny)}}

Problem 5: \frac{d}{dx}(\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}}) is equal to:

(a) 2\sin{(2x)} (b) \sin{(2x)} (c) -2 \sin{(2x)} (d) 2\cos{(2x)}

Problem 6: If y=\log_{5}{(\log_{5}{x})} then the value of \frac{dy}{dx} is

(a) \frac{1}{x \log_{5}{x}} (b) \frac{1}{x \log_{5}{x}. (\log{5})^{2}} (c) \frac{1}{\log{5}.x\log{x}} (d) \frac{1}{x(\log_{5}{x})^{2}}

Problem 7: \frac{d}{dx}(ax+b)^{cx+d} is equal to:

(a) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + \log{(ax+b)}) (b) (ax+b)^{cx+d}(\frac{cx+d}{ax+b} + c \log{(ax+b)}) (c) a(ax+b)^{cx+d} (d) none

Problem 8: \frac{d}{dx}(\log{(\frac{\sin{(x-b)}}{\sin{(x-a)}})}) is equal to:

(a) \frac{\cos{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (b) \frac{\sin{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

(c) \frac{\sin{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (d) \frac{\cos{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}

Problem 9: If y = \sqrt{\frac{\sec{x}+\tan{x}}{\sec{x}-\tan{x}}} and 0<x<\frac{\pi}{2}, then \frac{dy}{dx} is :

(a) \sec{x}(\sec{x}-\tan{x}) (b) \sec{x}(\sec{x}+\tan{x}) (c) \tan{x}(\sec{x}+\tan{x}) (d) \tan{x}(\sec{x}-\tan{x})

Problem 10: \frac{d}{dx}e^{ax}(a\sin{(bx)}-b\cos{(bx)}) is equal to:

(a) e^{ax}(\sin{(bx)}) (b) (a^{2}+b^{2})e^{ax}\sin{(bx)} (c) e^{ax}\cos{(bx)} (d) (a^{2}+b^{2})e^{ax}\cos{(bx)}

Cheers,

Nalin Pithwa.

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