Derivatives part 14: IITJEE maths tutorial problems for practice

This is part 14 of the series

Question 1:

Let f(x)= \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}} for x<26 be a real valued function. Then, find f^{'}(x) for 1<x<26:

Answer 1:

Consider (\sqrt{x-1}-5)^{2} = x-1+25-10\sqrt{x-1} = x+24 -10\sqrt{x-1} so that we have


Hence, f(x) = \sqrt{x-1} + \sqrt{x-1}-5 = 2\sqrt{x-1}-5 when 1<x<26

Hence, f^{'}{x} = \frac{-2}{2\sqrt{x-1}} = -\frac{1}{\sqrt{x-1}}

Question 2:

Let 3f(x) - 2 f(\frac{1}{x})=x, then find f^{'}(2).

Answer 2:

Given that 3f(x) - 2 f(\frac{1}{x})=x….call this I.

Also, from above, we get 3f(\frac{1}{x}) - 2 f(x)= \frac{1}{x}…call this II.

so we get 6f(x)-4f(\frac{1}{x})=2x….call this I’

and 9f(\frac{1}{x})-6f(x) = 9/x…call this II’.

5f(\frac{1}{x})=2x+ \frac{9}{x} and hence, f^{'}(1/x) = \frac{2x}{5} + \frac{9}{5x}

Also, again 3f(x)-2f(1/x)=x….A


So, we now we get the following two equations:



so, now we have 5f(x) = 3x + \frac{2}{x} so that we get f(x) = \frac{3}{5}x+\frac{2}{5x} andf(1/x) = \frac{2}{5}x+\frac{9}{5x}

so f^{'}(x) = \frac{3}{5} + \frac{2}{5}\frac{-1}{x^{2}} = \frac{3}{5} - \frac{1}{10}=\frac{1}{2}

Question 3:

If x= \frac{a(1-t^{2})}{1+t^{2}} and y = \frac{2bt}{1+t^{2}}, then find \frac{dy}{dx}.

Answer 3:

Given that x = \frac{a(1-t^{2})}{1+t^{2}} where a is a parameter (constant) and t is a variable.

Let t=\tan{\theta} so that x = \frac{a(1-\tan^{2}{\theta})}{1+\tan^{2}{\theta}} = a \cos{2\theta}

so that y = \frac{2bt}{1+t^{2}}=b \sin{2\theta}

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2b\cos{2\theta}}{-2a\sin{2\theta}}=- \frac{b}{a\tan{2\theta}}

so that we have

\frac{dy}{dx} = - \frac{b(1-t^{2})}{a}

Question 4:

If y = \arccos{\frac{x-x^{-1}}{x+x^{-1}}} then find \frac{dy}{dx}

Answer 4:

Given that \arccos{\frac{x^{2}-1}{x^{2}+1}} = \arccos{\frac{1-x^{2}}{1+x^{2}}} and put x=\tan{\theta}

\frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}}=\cos{2\theta} so that y = \arccos {\cos{2\theta}}=2\theta

\frac{dy}{dx} = 2\frac{d}{dx}(\arctan{x})=\frac{2}{1+x^{2}} which is the required answer.

Question 5:

If \arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}=a, where a is a parameter, then find \frac{dy}{dx}.

Answer 5:

Given that a=\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})} so that \sin{a} = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}

(x^{2}+y^{2})\sin{a} = x^{2}-y^{2}

Differentiating both sides w.r.t. x, we get

2x\sin{a} + \sin{a} ,2y.\frac{dy}{dx}=2x-2y\frac{dy}{dx}


\frac{dy}{dx} = \frac{2x(1-\sin{a})}{1+\sin{a}}=\frac{x}{y}\times \frac{2y^{2}}{2x^{2}}=\frac{y}{x}

Question 6:

If y = cot^{-1}{(\sqrt{\frac{1+x}{1-x}})} then find \frac{dy}{d(\arccos{x})}.

Answer 6:

Given that y = cot^{-1}(\sqrt{(\frac{1+x}{1-x})}) so that y = \arctan{(\sqrt{(\frac{1-x}{1+x})})} = \arctan{(cot {(2\theta)})} where x=\tan^{2}{\theta} so that \frac{d\theta}{dx} = \frac{1}{1+x^{2}}

and \sec^{2}{y}.\frac{dy}{dx} = - cosec^{2}{(2\theta)}.2\frac{d\theta}{dx}

Let f=\arccos{x} so that \frac{df}{dx} = - \frac{1}{\sqrt{1-x^{2}}}

Now, note that \sec^{2}{y} = cosec^{2}{2\theta} so we get the following simplification:

\frac{dy}{dx} = - \frac{2}{1+x^{2}}

Now, \frac{dy}{df} = -\frac{\frac{dy}{dx}}{\frac{df}{dx}}= \frac{2\sqrt{1-x^{2}}}{1+x^{2}}


Nalin Pithwa

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