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KEAM 2010 Math Paper

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Question : 30 of 120

Marks: +1, -0

The derivative of

\sin^{-1}(2x\sqrt{1-x^{2}})

with respect to

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

is

$\; \frac{2}{3}$
$\; \frac{3}{2}$
$\; \frac{1}{2}$
1
0

Solution:

Let

y=\sin^{-1}(2x\sqrt{1-x^{2}})

.....(i)

and

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

....(ii)

Now, putting

x=\cos\theta

in Eq. (i), we get

y=\sin^{-1}(2\cos\theta\sqrt{1-\cos^{2}\theta})

=\sin^{-1}(2\cos\theta\sin\theta)

=\sin^{-1}(\sin 2\theta)

\Rightarrow\; y=2\theta

\Rightarrow\; y=2\cos^{-1}x

Differentiating it w.r.t.

\theta,

we get

\; \frac{dy}{d\theta}=2

...(iii)

Also, putting

x=\sin\theta

in Eq. (ii), we get

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

\therefore\; z=3\theta

Differentiating it w.r.t.

\theta

, we get

\frac{dz}{d\theta}=3

...(iv)

Now,

\; \frac{dy}{dz} = \; \frac{dy}{d\theta} \cdot \; \frac{d\theta}{dz}

=2 \cdot \; \frac{1}{3} = \; \frac{2}{3}

\therefore\; \frac{d(\sin^{-1} 2x\sqrt{1-x^{2}})}{d(\sin^{-1}(3x-4x^{3}))} = \; \frac{2}{3}

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