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)

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.

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