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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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