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MHTCET 2016 Math Solved Paper

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Question : 2 of 50

Marks: +1, -0

Derivative of

\tan^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)

with respect to

\sin^{-1}(3x-4x^{3})

is

$\frac{1}{\sqrt{1-x^{2}}}$
$\frac{3}{\sqrt{1-x^{2}}}$
3
$\frac{1}{3}$

Solution:

(d) Let

u = \tan^{-1} \left(\frac{x}{\sqrt{1-x^{2}}}\right)

and

v = \sin^{-1} (3x-4x^{3})

Now, put x = sin θ ⇒ θ =

\sin^{-1}(x),

then

u = \tan^{-1} \left(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\right)

and

v = \sin^{-1} (3\sin\theta-4\sin^{3}\theta)

⇒

u = \tan^{-1} \left(\frac{\sin\theta}{\cos\theta}\right)

and

v = \sin^{-1}(\sin 3\theta)

u = \tan^{-1}(\tan\theta)

and

v = \sin^{-1}(\sin 3\theta)

u = \theta

and

v = 3\theta

u = \sin^{-1} x

and

v = 3\sin^{-1} x

On differentiating both sides w.r.t. x, we get

\frac{du}{dx} = \frac{1}{\sqrt{1-x^{2}}}

and

\frac{dv}{dx} = 3 \times \frac{1}{\sqrt{1-x^{2}}}

∴

\frac{du}{dv} = \frac{ \frac{du}{dx} }{ \frac{dv}{dx} }

=

\frac{ \frac{1}{\sqrt{1-x^{2}}} }{ \frac{3}{\sqrt{1-x^{2}}} } = \frac{1}{3}

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