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UPSC NDA Math Model Paper 3

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Question : 59 of 120

Marks: +1, -0

What is the slope of the tangent to the curve

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

x=0?

0
1
2
None of the above

Solution:

Given:

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

To find: Slope of tangent at x = 0

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

Differentiating to both sides w.r.t x we get,

\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-\sin^{4}x}} \times \frac{d(\sin^{2}x)}{dx}

\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-\sin^{2}x}} \times 2 \sin x \times \cos x

So, calculating

\frac{dy}{dx}

at

x=0.

\frac{dy}{dx_{x=0}}=\frac{1}{\sqrt{1-0}} \times 2 \times 0 \times 1

\Rightarrow \frac{dy}{dx_{x=0}}-0

Hence, the slope of the tangent to the given curve is 0.

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