 In trigonometry we define the inverse trigonometric functions, the inverse sine, also known as arc sine, the inverse tangent, also known as arc tangent, the inverse secant, also known as arc secant, and while there are inverse, cosine, cotangent and cosecant functions, nobody cares about these. In calculus we'll find their derivatives. So suppose y is the inverse sine of x. So definitions are the whole of mathematics, all else is commentary. Our definition for inverse sine of x is, so that says sine y equals x. We can use implicit differentiation. Since we originally had y as a function of x, we should express the derivative as a function of x. So from the Pythagorean identity we have, let x equal sine y, and we do have to choose which square root. So again, definitions are the whole of mathematics, all else is commentary. Our definition of the inverse sine says that y is between negative pi halves and pi halves. And since y is in that interval, cosine y is non-negative, so we choose the positive square root. And so this gives us a formula for the derivative of arc sine. And we can find the derivative the normal way. We'll ignore everything but the last function. The derivative of arc sine is 1 over square root whatever squared times the derivative, put things back where you found them, add complete the derivative, and whatever simplifications you care to make. Similarly, if y equals arc tangent of x, then tangent of y equals x, and implicit differentiation gives us, we can use our Pythagorean identity, and since tangent of y equals x, and so we have the derivative of arc tangent. And all the usual rules apply. So this is arc tangent of whatever, so we differentiate, put things back where we found them, and simplify. The derivative of arc secant is a little more complicated. Here we'll have to pay close attention to the definition. We start the same way. If y equals arc secant x, then secant y equals x, we'll use implicit differentiation. From the Pythagorean identity, and secant y equals x, we have, and here's the problem. If y is between zero and pi halves, tangent y is positive, so we take the positive root. But if y is between pi halves and pi, tangent y is negative, so we take the negative root. And so our derivative for arc secant has two possible values. Or does it? Well, no, there's only one possible derivative, and the reason is this. While we could write our derivative this way, notice that for values between zero and pi halves, both secant y and tangent y are positive, and to the interval between pi halves and pi, both secant y and tangent y are negative. So this product, secant y, tangent y, is always positive. And since the square root is positive, we can guarantee the product is always positive by using the absolute value. And so we can write our derivative of arc secant as... So if I want to find the derivative of arc secant of e to the x, we get... And since e to the x is positive for all values of x, then the absolute value of e to the x is e to the x, and we can remove the common factor to get...