Understanding d/dx (sin-1x) & Deriving dy/dx = -sin-2xcosx

[Kyoma]

d/dx (sin^-1x) = \frac{1}{\sqrt{1-x²}}

But using chain rule, I got:

dy/dx = -sin^-2xcosx

Why?

 

[gb7nash]

d/dx (sin^-1x) = \frac{1}{\sqrt{1-x²}}

But using chain rule, I got:

dy/dx = -sin^-2xcosx

Why?

sin^-1x is not equal to \frac{1}{\sin x}. sin^-1x stands for the arcsin of x, or more specifically, the angle y that satisfies sin y = x.

I've always personally disliked the notation sin^-1x, but it is what it is.

 

[HallsofIvy]

If y= sin^{-1}(x) then x= sin(y).

\frac{dx}{dy}= cos(y)

so that
\frac{dy}{dx}= \frac{1}{cos(y)}

but y= sin^{-1}(x) so cos(sin^{-1}(x))= \sqrt{1- sin^2(sin^{-1}(x))}= \sqrt{1- x^2}

so
\frac{d(sin^{-1}(x)}{dx}= \frac{1}{\sqrt{1- x^2}}

As gb7nash said, the "-1", applied to functions denotes the "inverse function" (the inverse for function composition), not the reciprocal (the inverse for multiplication).