# Inverse of sinx with arcsinx?

• Jun 25th 2008, 02:49 PM
SportfreundeKeaneKent
Inverse 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, 02:59 PM
Moo
Hello,

Quote:

Originally Posted by SportfreundeKeaneKent
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

arcsin x denotes the inverse function, as stated above.

This means :

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

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

I'll do it the other way round :

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

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

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

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

Any question ?