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## Logarithmic functions differentiation

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# Worked example: Derivative of log₄(x²+x) using the chain rule

AP Calc: FUN‑3 (EU), FUN‑3.C (LO), FUN‑3.C.1 (EK)

## Video transcript

- [Voiceover] Let's say that Y is equal to log base four of X squared plus 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 four, not just of X, but we're taking that
of another expression that involves X. So we could say 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 squared plus X. And it's gonna be useful later on to know what U prime of X is. So that's gonna be I'm just gonna use the power rule here so two X plus one I brought that two out front and decremented the exponent. Derivative with respect to X of X is one. And we can say the log base four of this stuff well we could call that a function V. We can say V of well if we said V of X this would be log base four of X. And then we've shown in other videos that V prime of X is, we're gonna be very
similar that if this was log base E, or natural log, except we're going to scale it. So it's going to be one over one over log base four. Sorry, one over the natural log. The natural log of four times X. If this was V of X, if V of X was just natural log of X, our
derivative would be one over X. But since it's log base four 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 four. You think of scaling the whole expression by one over the natural log of four. 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 four 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 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. 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 know what 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 X, U of X. And you just do is you take the derivative of the green function with
respect to the blue function. So it's going to be one over the natural log of four. The natural log of four. Times, instead of putting an X there it would be times U of X. 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 one over the natural log of four. U of X is X squared plus X. So X squared plus X. And then we're gonna multiply that times U prime of X. So times two X plus one. And so we can just rewrite this as two X plus one over over over the natural log of four. The natural log of four times X squared plus X. Times X squared plus X. And we're done, 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.