## 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:

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

$\sqrt{x+24-10\sqrt{x-1}}=\sqrt{x-1}-5$

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

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 $f^{'}(2)$.

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

$3f(1/x)-2f(x)=1/x$…B

So, we now we get the following two equations:

$9f(x)-6f(1/x)=3x$…..A’

$6f(1/x)-4f(x)=2/x$….B’

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

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 $\frac{dy}{dx}$.

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 $\frac{dy}{dx}$

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 $\frac{dy}{dx}$.

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

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

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

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

Question 6:

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

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

Cheers,

Nalin Pithwa

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