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

Let's say that Y is equal to log base 4 of x^2 + x. What is the derivative of y with respect to x going to be equal to? Now, you might recognize immediately that this is a composite function. We're taking the log base 4 not just of x, but we're taking that of another expression that involves x.

So we could say, this thing in blue, that's U of x. Let me do that in blue. So this thing in blue that is U of x, U of x is equal to x^2 + x. It's going to be useful later on to know what U prime of x is. So that's going to be just going to use the power rule here, so 2x + 1.

BR that brought that two out front and decremented the exponent. The derivative with respect to x of x is 1. We could say the log base 4 of this stuff, well, we could call that a function V. We could say V of, well, if we said V of x, this would be log base 4 of x.

And then we've shown in other videos that V prime of x is going to be very similar if this was log base e or natural log, except we're going to scale it. So it's going to be 1 over the natural log of 4 times x. If this was V of, if V of x was just natural log of x, our derivative would be 1/x.

But since it's log base 4, and this comes straight out of the change of base formulas that you might have seen, and we have a video where we show this, but we just scale it in the denominator with this natural log of 4. Or you could think of scaling the whole expression by 1 over the natural log of 4.

But we can now use this information because y, this y can be viewed as V of V of, remember, V is the log base 4 of something. But it's not V of x; we don't have just an x here. We have the whole expression that defines U of x. We have U of x right there.

And let me draw a little line here so that we don't get those two sides confused. And so we know from the chain rule the derivative of y with respect to x. This is going to be, this is going to be the derivative of V with respect to U. Or we could call that V prime, V prime of U of x, let me do the U of x in blue, V prime of U of x times U prime of x.

Well, what is V prime of U of x? We know what V prime of x is. If we want to do V prime of U of x, we would just replace wherever we see an x with a U of x. So this is going to be equal to V prime of U of x, and you just view it as you're taking the derivative of the green function with respect to the blue function.

So it's going to be 1 over the natural log of 4 times U of x. And of course, that whole thing times U prime of x. And so, and I'm doing more steps just hopefully so it's clearer what I'm doing here. So this is 1 over the natural log of 4 times U of x is x^2 + x, so x^2 + x.

And we're going to multiply that times U prime of x, so times 2x + 1. So we can just rewrite this as (2x + 1) / (natural log of 4) * (x^2 + x). And we could distribute this natural log of four if we found that interesting, but we have just found the derivative of y with respect to x.