Derivatives: part 6: IITJEE tutorial practice problems

Problem 1:

If \sec {(\frac{x+y}{x-y})}=a, then \frac{dy}{dx} is (i) \frac{x}{y} (ii) \frac{y}{x} (iii) y (iv) x

Problem 2:

If f(x) = x+ 2, when -1<x<1;

f(x)=5, when x=3;

f(x) = 8-x, when x>3; then, at x=3, the value of f^{'}(x) is

(a) 1 (b) -1 (c) 0 (d) does not exist.

Problem 3:

If y = x \tan{y}, then \frac{dy}{dx} is equal to

(i) \frac{\tan{y}}{x-x^{2}-y^{2}} (ii) \frac{\tan{y}}{y-x}

(iii) \frac{y}{x-x^{2}-y^{2}} (iv) \frac{\tan{x}}{x-y^{2}}

Problem 4:

If g is the inverse function of f and f^{'}(x) = \frac{1}{1+x^{n}}, then g^{'}(x) is equal to

(i) 1 + (g(x))^{n} (ii) 1+g(x) (iii) 1-g(x) (iv) 1-(g(x))^{n}

Problem 5:

If f(x) = \log_{x^{2}}(\log{x}) then f(x) at x=c is :

(i) 0 (ii) 1 (iii) \frac{1}{e} (iv) \frac{1}{2e}

Problem 6:

If y = (\sin{x})^{\tan{x}} then \frac{dy}{dx} is equal to :

(i) (\sin{x})^{\tan{x}}(1+ \sec^{2}{x} \log{\sin{x}})

(ii) \tan{x}. (\sin{x})^{\tan{x}-1} \times \cos{x}

(iii) (\sin{x})^{\tan{x}}\sec^{2}{x} \times \log{\sin{x}}

(iv) \tan{x} (\sin{x})^{\tan{x}-1}

Problem 7:

If y = \sqrt{\sin{x}+y}, then \frac{dy}{dx} equals:

(i) \frac{\sin{x}}{2y-1} (ii) \frac{\sin{x}}{1-2y} (iii) \frac{\cos{x}}{1-2y}

(iv) \frac{\cos{x}}{2y-1}

Problem 8:

If x = \sqrt{\frac{1-t^{2}}{1+t^{2}}} and y = \sqrt{\frac{\sqrt{1+t^{2}}-sqrt{1-t^{2}}}{\sqrt{1+t^{2}}+\sqrt{1-t^{2}}}}

then the value of \frac{d^{2}y}{dx^{2}} at t=0 is given by:

(a) 0 (b) 1/2 (c) 1 (d) -1

Problem 9:

If x = a \cos^{3}{\theta}, y = a \sin^{3}{\theta}, then \sqrt{1 + (\frac{dy}{dx})^{2}} is equal to:

(i) \sec^{2}{\theta} (ii) \tan^{2}{\theta} (iii) \sec{\theta} (iv) |\sec{\theta}|

Problem 10:

If y = \arcsin{\sqrt{1-x}} + \arccos{\sqrt{x}}, then \frac{dy}{dx} equals:

(a) \frac{1}{\sqrt{x(1-x)}} (b) \frac{1}{x(1+x)} (c) \frac{-1}{\sqrt{x(1-x)}} (d) none

Regards,

Nalin Pithwa

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