 An important set of derivatives are those of the trigonometric and the inverse trigonometric functions. Now we can generally find these using a little bit of trigonometry and a little bit of algebra. So for example, let's say I want to find the derivative of tangent of x. Now we can use our definition of tangent as sine over cosine. So when we differentiate, remember that the type of function is determined by the last operation performed. In this case, the last thing we do is divide so our derivative will begin by using the quotient rule. So we'll get cosine times derivative sine minus sine times derivative cosine all over cosine squared. But we have some unresolved derivatives that of sine and cosine. Fortunately we know what those are, which will give us one form of the derivative of tangent. However, this forms a little bit messy, so maybe we'll do a little bit of algebra. And do some trigonometry. Our numerator cosine squared x plus sine squared x is equal to one. So our derivative of tangent is one over cosine squared x. But we can do a little bit more trigonometry. Since one of our cosine is secant, then our derivative can be written as secant squared x. And we'll go ahead and record our process. We started with the definition of tangent as sine over cosine. Then we applied the quotient rule, a little bit of algebra, and a little bit of trigonometry to get the derivative of tangent equal to secant squared. And we can find the derivatives of the other trigonometric functions this way, but we won't bother. Because if we ever need to know the derivative of secant or the derivative of cosecant or the derivative of cotangent, we can figure them out by knowing what the function is in terms of sine and cosine, knowing the basic derivative rules, knowing the derivatives of sine and cosine, and being able to do a little bit of algebra and a little bit of trigonometry. How about the inverse trigonometric functions? Say I want to find the derivative of y equals arctangent of x. So again, the key idea here is that the definition of arctangent is that if y equals arctangent of x, then tangent of y is equal to x. So let's use implicit differentiation. So on the left-hand side I have tangent something, and the derivative of tangent is secant squared times the derivative of something. Everything back where we found it gives me the left-hand side. Meanwhile, on the right-hand side, the derivative of x is just 1. Now I'll solve for dy dx, which gives me 1 over secant squared y. And an important rule is since we originally had a formula y in terms of x, our final answer should also be a formula in terms of x. And so to simplify this expression 1 over secant squared y, I'm going to use a little trigonometry. The Pythagorean identity says that tangent squared y plus 1 is equal to secant squared y. And so this 1 over secant squared y becomes 1 over tangent squared y plus 1, but I can now reduce that because I know the tangent of y is equal to x. And so this gives me... So putting these ideas together from tangent y equals x, and the Pythagorean identity tangent squared y plus 1 equals secant squared, we find the derivative of arc tangent is 1 over secant squared, which is 1 over x squared plus 1.