# If for     , the derivative of    is   then equals : Option 1) Option 2) Option 3) Option 4)

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)

This option is incorrect.

Option 2)

This option is incorrect.

Option 3)

This option is incorrect.

Option 4)

This option is correct.

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