Exponential functions differentiation | Advanced derivatives | AP Calculus AB | Khan Academy

Let's say that Y is equal to 7/2 the x squared minus X power. What is the derivative of Y, derivative of Y with respect to X? And like always, pause this video and see if you can figure it out.

Well, based on how this has been color-coded ahead of time, you might immediately recognize that this is a composite function. It could be viewed as a composite function if you had a V of X, which is equal to 7 to the X power.

And you had another function U of X, U of X, which is equal to x squared minus X. Then what we have right over here is Y. Y is equal to 7 to something, so it's equal to V of... and it's not just V of X, it's V of U of X. Instead of an X here, you have the whole function U of X, x squared minus X.

So it's V of U of X, and the chain rule tells us that the derivative of Y with respect to X... and you'll see different notations here. Sometimes you'll see it written as the derivative of V with respect to U, so V prime of U of X times the derivative of U with respect to X.

So that's one way you could do it, or you could say that this is equal to... this is equal to the derivative, the derivative of V with respect to X, derivative... or sorry, derivative of V with respect to U, DV/DU times the derivative of U with respect to X.

And so either way, we can apply that right over here. So what's the derivative of V with respect to U? What is V prime of U of X? Well, we know... let me actually write it right over here. If V of X is equal to 7 to the X power, V prime of X would be equal to... and we prove this in other videos where we take derivatives of exponentials of bases other than e.

This is going to be the log of 7 times 7 to the X power. So if we are taking V prime of U of X, then notice instead of an X everywhere, we're going to have a U of X everywhere.

So this right over here is going to be the natural log of 7 times 7 to the... instead of saying 7 to the X power, remember we're taking V prime of U of X, so it's going to be 7 to the x squared minus x power.

Then we want to multiply that times the derivative of U with respect to X, so U prime of X... well, that's going to be 2x to the 1st, which is just 2x minus 1.

So we're going to multiply this times 2x, 2x minus 1. So there you have it, that is the derivative of Y with respect to X.

We could try to simplify this or I guess re-express it in different ways, but the main thing to realize is, look, we're just going to take the derivative of the 7 to the U of X power with respect to U of X.

So we treat the U of X the way that we would have treated an X right over here. So it's going to be the natural log of 7 times 7 to the U of X power; we take that and multiply that times U prime of X.

And once again, this is just an application of the chain rule.