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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

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 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 plus 1 drawing that brought that 2 out front and decremented the exponent derivative with respect to X of X is is 1 and 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 that if this was log base E or natural log except we're going to scale it so it's going to be 1 over 1 over log base 4 sorry 1 over the natural log the natural log of 4 times X if this was V Evette if the V of X was just natural log of X our derivative be 1 over X but since its log base 4 and this comes straight of the change of base formula is 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 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 be prime of U of X we would just replace wherever we see an X with the U of X so this is going to be equal to V prime of u at U of X and you just use 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 natural log of 4 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 clear what I'm doing here so this is 1 over the natural log of 4 U of X is x squared plus X so x squared plus X and we're going to multiply that times u prime of X so times 2x plus 1 and so we can just rewrite this as 2x plus 1 over over over the natural log of 4 natural log of 4 times x squared plus x times x squared plus X and we're done we could distribute this natural log of 4 if we found that interesting but we have just found the derivative of Y with respect to X
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