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
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 ?