I don't even understand the wording of this. Also arcSINx?

Denote by y=arcsinx,-1< or =x< or =1, the inverse of y=sinx with x between -pi/2 and pi/2. Show that

d/dx arcsinx = 1/sqrt(1-x^2)with x between -1 and 1

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- Jun 25th 2008, 01:49 PMSportfreundeKeaneKentInverse of sinx with arcsinx?
I don't even understand the wording of this. Also arcSINx?

Denote by y=arcsinx,**-1**< or =**x**< or =**1**, the inverse of y=sinx with x between -pi/2 and pi/2. Show that

**d/dx arcsinx = 1/sqrt(1-x^2)**with x between -1 and 1 - Jun 25th 2008, 01:59 PMMoo
Hello,

arcsin x denotes the inverse function, as stated above.

This means :

$\displaystyle y=\arcsin x \Longleftrightarrow x=\sin y$.

You want $\displaystyle \frac{dy}{dx}$.

I'll do it the other way round :

$\displaystyle \frac{dx}{dy}=\frac{d}{dy}(\sin y)=\cos y=\pm \sqrt{1-\sin^2 y}$

Here, $\displaystyle \cos y={\color{red}+}\sqrt{1-\sin^2 y}$, because -pi/2<=y<=pi/2 --> cos y is positive.

$\displaystyle \frac{dx}{dy}=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$, because $\displaystyle x=\sin y$.

Therefore $\displaystyle \frac{d}{dx} \arcsin x=\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

Any question ?