So there is a question that asks that to show that sinx has an inverse function, and show that it;s differantiable at any point. So I showed that it's one to one and it's bijection func at [-pie/2,pie/2], and ther3fore exists an inverse function in that interval. However I am having a trouble showing that it's differantiable. I had an idea to show that 1/cosx is defined in (-pie/2,pie/2) and therefore according to the inverse derivative it's differantiable at any point in that interval if there is a derivative of that function. However, I think there is a problem with my logic, because according to what I did ,I think , I already assumed that it's differantiable.
The other thing that I did , I showed that the inverse function is continuous at all points and it's an increasing function. So, in my point of view, every function that is increasing and it's continuous , is differantiable. However I couldn't prove that.
I tried to play with the definition of differentiation and somehow get something . however I failed. Can someone give me a hint. Please note that , at this point they assume that we know nothing about arcsin. so if there is a trigonometric identity, I cant use that.