Derivatives of tan(x) and cot(x) | Derivative rules | AP Calculus AB | Khan Academy

We already know the derivatives of sine and cosine. We know that the derivative with respect to x of sine of x is equal to cosine of x. We know that the derivative with respect to x of cosine of x is equal to negative sine of x.

So, what we want to do in this video is find the derivatives of the other basic trig functions. In particular, we know, let's figure out what the derivative with respect to x is. Let's first do tangent of x. Tangent of x, well, this is the same thing as trying to find the derivative with respect to x of tangent of x. Tangent of x is just sine of x over cosine of x.

Since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x. So, times sine of x minus the top function, which is sine of x, times the derivative of the bottom function.

So, the derivative of cosine of x is negative sine of x. So, I could put the sine of x there, but where the negative can just cancel that out and it's going to be over the bottom function squared, so cosine squared of x. Now, what is this?

Well, what we have here, this is just cosine squared of x. This is just sine squared of x. We know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x is going to be equal to one for any x. So, all of this is equal to 1, and so we end up with 1 over sine squared of x, which is the same thing as secant squared of x.

So, this is just secant squared of x. So that’s pretty straightforward. Now, let's just do the inverse, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. So, that was fun, so let's do that.

The derivative with respect to x of cotangent of x, well, same idea; that's the derivative with respect to x. And this time, let me make some sufficiently large brackets. So, now this is cosine of x over sine of x. But once again we can use the quotient rule here.

So, this is going to be the derivative of the top function, which is negative sine of x times the bottom function, so times sine of x, minus the top function, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared, so sine squared of x.

Now, what does this simplify to? Let's see. This is sine of x, although we have a negative there, minus cosine squared of x. But we could factor out the negative, and this would be sine squared of x plus cosine squared of x. Well, this is just one by the Pythagorean identity.

And so, this is negative one over sine squared of x. Negative one over sine squared of x, and that is the same thing as negative cotangent squared of x. There you go.