Chain rule for derivative of log function

- [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.