Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If for $x\: \epsilon \left ( o,\frac{1}{4} \right )$ , the derivative of

$\tan ^{-1}\: \left ( \frac{6x\sqrt{x}}{1-9x^{3}} \right )$ is $\sqrt{x}\: .g(x)$ then $g(x)$ equals :

Option 1)
$\frac{3x\sqrt{x}}{1-9x^{3}}$

Option 2)
$\frac{3x}{1-9x^{3}}$

Option 3)
$\frac{3}{1+9x^{3}}$

Option 4)
$\frac{9}{1+9x^{3}}$

Answers (1)

best_answer

As we learnt in

Inverse Trigonometric Functions -

The functions $\sin ^{-1}x, \cos ^{-1}x, \tan ^{-1}x, \cot ^{-1}x, \csc ^{-1}x$ and $\sec ^{-1}x$ are the inverse trigonometric functions.

- wherein

If $\sin x = \frac{1}{2}$

then, $x = \sin ^{-1}\frac{1}{2}$

Let, $\tan^{-1}\:\frac{6x\sqrt x}{1-9x^{3}}=\Theta$

$\tan \Theta=\frac{6x\sqrt x}{1-9x^{3}}$

Differentiating both sidesw.r.t. x we get:

$\frac {d \Theta}{d x}\sin^{2}\Theta=\frac{(1-9x^{3})\:9\sqrt x-6x^{\frac{3}{2}}(-27x^{2})}{(1-9x^{3})^{2}}$

$\frac {d \Theta}{d x}\sin^{2}\Theta=9\sqrt x\:\: \frac{(1+9x^{3})}{(1-9x^{3})^{2}}$ ................ ( 1 )

Now, $\tan \Theta=\frac{6x\sqrt x}{1-9x^{2}}\:\:\:\Rightarrow \sin^{2}\Theta = 1+\frac{(6x\sqrt x)^{2}}{(1-9x^{3})^{2}}$

$=1+\frac{36x^{3}}{(1-9x^{3})^{2}}$

$=\frac{1+81x^{6}+18x^{3}}{(1-9x^{3})^{2}}$

$\sec^{2}\Theta=\frac{(1+9x^{3})^{2}}{(1-9x^{3})^{2}}$ ................... ( 2 )

From ( 1 ) and ( 2 )

$\frac{d \Theta}{dx}=\frac{9\sqrt x (1+9x^{3})^{2}}{(1-9x^{3})^{2}}\:\:\frac{(1-9x^{3})^{2}}{(1+9x^{3})^{2}}$

$\frac{d \Theta}{dx}=\frac{9\sqrt x}{1+9x^{2}}$

On comparison, $g(x)=\frac{9}{1+9x^{3}}$

Option 1)

$\frac{3x\sqrt{x}}{1-9x^{3}}$

This option is incorrect.

Option 2)

$\frac{3x}{1-9x^{3}}$

This option is incorrect.

Option 3)

$\frac{3}{1+9x^{3}}$

This option is incorrect.

Option 4)

$\frac{9}{1+9x^{3}}$

This option is correct.

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA 2024 round 4 seat allotment results out at josaa.nic.in; report online by July 15

Latest Question