Question: $$ \dfrac{d}{dx} \sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$$

$$\dfrac{d}{dx} \cos^{-1}x = - \dfrac{d}{dx} \sin^{-1}x$$

What is the reason for this?

Answer: 
Because $\cos^{-1} x=\frac{\pi}{2}-\sin^{-1} x$.
Try graphing the $\sin$ and $\cos$ functions. Notice how $\sin$ (the blue wave) is just a reflection of $\cos$ (the red wave) about $\theta=\pi/4$ (in the first quadrant of the unit circle, where $\sin\theta=\cos\theta$). If you were to draw a horizontal line, what would the derivatives at the points of intersection of the horizontal line and the red and blue waves be (moving outwards from $\theta=\pi/4$)? It should be clear why the derivatives of the inverses of the $\sin$ and $\cos$ functions are additive inverses of each other.

![enter image description here](https://i.stack.imgur.com/BpjGz.png)