- #1

Saitama

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## Homework Statement

Find [itex]\frac{dy}{dx}[/itex].

[tex]y=\sin^{-1}(2x\sqrt{1-x^2}), \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}[/tex]

## Homework Equations

## The Attempt at a Solution

I started with substituting x=sinθ.

The expression simplifies to [itex]y=\sin^{-1}(\sin(2θ))[/itex] which is equal to [itex]y=2θ[/itex].

Substituting back the value of θ, i get [itex]y=2\sin^{-1}x[/itex].

Therefore [itex]\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}[/itex]

According to the book, this answer is correct.

But if a start by substituting x=cosθ. The expression simplifies to [itex]y=2\cos^{-1}θ[/itex], if i differentiate this, i get

[itex]\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}[/itex]

I don't understand why i get these two different answers, i suppose it has to do something with the range of x given in the question but i don't seem to get the point.

Any help is appreciated!

Thanks!

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