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

http://www.math.brown.edu/UTRA/trigderivs/image017.gif. Can someone help me understand please!

Thank you!

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- Jan 27th 2009, 08:43 AMgabet16941derivative of arcsine
I dont understand why when the derivative of arcsine is taken it is

http://www.math.brown.edu/UTRA/trigderivs/image017.gif. Can someone help me understand please!

Thank you! - Jan 27th 2009, 09:30 AMChris L T521
Note that $\displaystyle \sin^{-1}x$ is differentiable on the interval $\displaystyle -1<x<1$ and $\displaystyle x=\sin y$ is differentiable on the interval $\displaystyle -\frac{\pi}{2}<y<\frac{\pi}{2}$

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

Now, implicitly differentiating both sides, we have

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

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

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

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

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

Does this make sense? - Jan 27th 2009, 09:37 AMgabet16941
Yes thank you very much!