Differential calculus is not the easiest course to study. You'll have to know many differentiation rules and how they are applied to very specific functions. Some of the most difficult derivatives to take are the trigonometric derivatives. For most of these derivatives you will simply want to memorize the answers. Going through the proofs of how these derivatives are found can be difficult at times, especially if you approach them from the fundamental concepts. Today we're going to take a look at the derivative of tan. We will see that deriving this derivative is not that difficult if we use existing knowledge of other trigonometric derivatives.

So the first thing to know is that the tan function is simply the sine function over the cosine function. By observing this simple fact we can see that we have the quotient function. This means that we can use the quotient rule to solve for the derivative of tan.

First off the prime of the sine function is simply equal to the cosine function. We know this from previous sections. Similarly, the prime of the cosine function is equal to the negative sine function. Now that we know these derivatives we can apply them in the quotient rule to solve for the derivative of tan.

y = sinx / cosx

y '= [cosx * cosx – (-sinx) * sinx] / (cosx) ^ 2

= [(Cosx) ^ 2 + (sinx) ^ 2] / (cosx) ^ 2

We know from Pythagorean theorem that the top part of this function is simply equal to one. We can simplify;

y '= 1 / (cosx) ^ 2 = (secx) ^ 2

So we can see now that the derivative of tan is simply equal to secant squared x. There are other more fundamental ways to construct this derivative, but this is the most common and easiest way. Try the other methods yourself and see if you get the same answers. Otherwise just memorize this equation as you will be using it a lot in the future.

Source by Carl Chute