Derivatives: part 7: IITJEE tutorial problems practice

Problem 1: Differential coefficient of \sec{\arctan{x}} is

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

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

(a) 2 (b) -2 (c) 1 (d) -1

Problem 3: If y = \arcsin{\sqrt{x-ax}-\sqrt{a-ax}}, then \frac{dy}{dx} is equal to:

(a) \frac{1}{2\sqrt{x}\sqrt{1-x}} (b) \sin{(\sqrt{x})} \times \sin{(\sqrt{a})}

(c) \frac{1}{\sin{\sqrt{a-ax}}} (d) zero

Problem 4: For the differentiable function f, the value of : \lim_{h \rightarrow 0} \frac{(f(x+h))^{2}-(f(x))^{2}}{2h} is equal to:

(a) (f^{'}(x))^{2} (b) \frac{1}{2}(f(x))^{2} (c) f(x)f^{'}(x) (d) zero

Problem 5: The derivative of \arctan{\frac{\sqrt{1+x^{2}}-1}{x}} w.r.t. \arctan{(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}})} at x=0 is :

(a) \frac{1}{8} (b) \frac{1}{4} (c) \frac{1}{2} (d) 1

Problem 6: If x = e^{y+e^{y+e^{y+e^{y+ \ldots}}}} then \frac{dy}{dx} is

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

Problem 7: Consider the following statements:

(1) (\frac{f}{g})^{'} = \frac{f^{'}}{g^{'}} (2) \frac{(fg)^{'}}{fg} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

(3) \frac{(f+g)^{'}}{f+g} = \frac{f^{'}}{f} + \frac{g^{'}}{g} (4) \frac{(f/g)^{'}}{f/g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}

Which of the following statements are true?

(a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4

Problem 8: If y=e^{x+3\log{x}} then \frac{dy}{dx} =

(a) e^{x+3\log{x}} (b) e^{x}.x^{2}(x+3) (c) e^{x}. e^{3\log{x}} (d) 3x^{2}e^{x}

Problem 9: If y=\sin^{2}(x \deg), then find the value of \frac{dy}{dx} is:

(a) \frac{\pi}{360}\sin{(2 x \deg)} (b) \frac{\pi}{2}\sin{(2x\deg)} (c) 180 \sin {(2x\deg)} (d) \frac{\pi}{180}\sin{(2x\deg)}

Problem 10: If y=\log_{a}{x} + \log_{x}{a} + \log_{x}{x}+ \log_{a}{a} then the value of \frac{dy}{dx} is:

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

Cheers,

Nalin Pithwa.

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