# Thread:

1. ## derivative of arcsine

I dont understand why when the derivative of arcsine is taken it is
. Can someone help me understand please!

Thank you!

2. Originally Posted by gabet16941
I dont understand why when the derivative of arcsine is taken it is
. Can someone help me understand please!

Thank you!
Note that $\sin^{-1}x$ is differentiable on the interval $-1<x<1$ and $x=\sin y$ is differentiable on the interval $-\frac{\pi}{2}<y<\frac{\pi}{2}$

Let $y=\sin^{-1}x\implies x=\sin y$.

Now, implicitly differentiating both sides, we have

$1=\cos y\frac{\,dy}{\,dx}\implies\frac{\,dy}{\,dx}=\frac{ 1}{\cos y}$

We can divide through by $\cos y$ because $\cos y\neq0$ on the interval $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Also $\cos y >0$ on the interval $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Now, we can use the identity $\cos^2y+\sin^2 y =1$ to find $\cos y$:

$\cos^2y+\sin^2y=1\implies \cos^2y=1-\sin^2y\implies \cos y=\sqrt{1-\sin^2y}$

Thus, we see that $\frac{\,dy}{\,dx}=\frac{1}{\sqrt{1-\sin^2y}}$

Since we let $x=\sin y$, it now implies that $\color{red}\boxed{\frac{\,d}{\,dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}}$

Does this make sense?

3. Yes thank you very much!
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