# Thread: Derivative of inverse sine?

1. ## Derivative of inverse sine?

I have been getting mixed signals about what the derivative of $arcsin(u)$ is from my math teacher. He says that $\frac{d}{du}[arcsin(u)] \: = \: \frac{du}{\sqrt{1 - u^{2}}}$, and that $\int \frac{du}{\sqrt{a^{2} + u^{2}}} \: = \: arcsin(\frac{u}{a})$, where $a$ is a constant.
I realized these are not the same derivative of $arcsin(u)$, so I checked Wolfram Alpha and it said that the integral version that I have been given actually solves for the inverse hyperbolic sine of u. Which is right?

2. ## Re: Derivative of inverse sine?

Originally Posted by j1234
I have been getting mixed signals about what the derivative of $arcsin(u)$ is from my math teacher. He says that $\frac{d}{du}[arcsin(u)] \: = \: \frac{du}{\sqrt{1 - u^{2}}}$, and that $\int \frac{du}{\sqrt{a^{2} + u^{2}}} \: = \: arcsin(\frac{u}{a})$, where $a$ is a constantt?
Look at this $\frac{d}{{du}}\left[ {\arcsin \left( {\frac{u}{a}} \right)} \right] = \frac{{\frac{1}{a}}}{{\sqrt {1 - \frac{{{u^2}}}{{{a^2}}}} }} = \frac{1}{{\sqrt {{a^2} - {u^2}} }}$

It is clear that a simple sign error was made in your instructor's example.