2

I have to use $\sin^{-1}(z) = -i\log(i(z + \sqrt{z^2 - 1}))$ to compute the derivative of $\sin^{-1}(z)$, $\frac{d}{dz}\sin^{-1}(z)$.

Here is my process: $$\sin^{-1}(z) = -i\log(i(z + \sqrt{z^2 -1}))$$ $$\sin(-i\log(i(z + \sqrt{z^2 - 1}))) = z$$ Differentiating both sides... $$\cos(-i\log(i(z + \sqrt{z^2 - 1}))) \cdot \left( 1 + \frac{1}{2}(z^2 - 1)^{-\frac{1}{2}} (2z)\right ) = 1$$ $$1 + \frac{z}{\sqrt{z^2 - 1}} = \frac{1}{ \cos(-i\log(i(z + \sqrt{z^2 - 1})))}$$ Since we are told that $\arcsin(z) = -i\log(i(z + \sqrt{z^2 - 1}))$, then we have the following: $$1 + \frac{z}{\sqrt{z^2 - 1}} = \frac{1}{ \sqrt{1 - z^2}} $$

Is there anything wrong with this?

Complex.Me
  • 21
  • You didn't get the derivative of $\arcsin{z}$? – KittyL Mar 03 '15 at 10:05
  • @KittyL I more so meant, why is it wrong. Shouldn't I have equality of both sides in the final equation? – Complex.Me Mar 03 '15 at 10:11
  • You forgot the $-i$ and the derivative of $ln...$ at the "differentiating both sides" step. But overall, this method wouldn't give you the derivative of $\arcsin{z}$. You should just use chain rule from the beginning. – KittyL Mar 03 '15 at 10:14
  • Why wouldn't this method work? Also, using the chainrule is way more logical than what I attempted. Wow. – Complex.Me Mar 03 '15 at 10:16
  • What you did already involve chain rule, but the answer does not have $\arcsin$, that's why it does not work. Also the problem ask you to use this way in order to avoid the inverse approach. – KittyL Mar 03 '15 at 10:19

0 Answers0