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).