derivative of arcsine

**gabet16941** · January 27th 2009, 08:43 AM

I dont understand why when the derivative of arcsine is taken it is

. Can someone help me understand please!

Thank you!

**Chris L T521** · January 27th 2009, 09:30 AM

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?

**gabet16941** · January 27th 2009, 09:37 AM

Yes thank you very much!

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