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**davidciprut** 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 [-pi/2,pi/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 (-pi/2,pi/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.