### Video Transcript

Find ππ¦ ππ₯, given that π¦
equals the natural logarithm of four π₯ plus five to the power of seven.

So the first thing that weβre
looking to do when we differentiate our function is actually rewrite it in another
way. And we can do that because of one
of the log laws. And that law states that if weβve
got log to the base π of π₯ to the power of π, then this can be rewritten as π
log to base π of π₯. So when we rewrite our function, we
actually get π¦ is equal to seven ln four π₯ plus five. So we did that using the log
law.

So now if weβre actually looking to
find ππ¦ ππ₯, we can say itβs gonna be equal to seven multiplied by πππ₯ of ln
four π₯ plus five. So what we wanna do now is actually
differentiate our ln four π₯ plus five. And the way to do that is actually
by using the chain rule. And the chain rule actually states
that ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. So we can actually adapt this and
use this to solve our problem.

So if weβre looking to
differentiate ln four π₯ plus five, we can actually say that π’ is gonna be equal to
four π₯ plus five, which is the actual expression inside the parentheses, and π¦ is
gonna be equal to ln π’. So first of all, what weβre
actually gonna do is differentiate π’. So itβs ππ’ ππ₯, which is just
gonna give us four, because if we differentiate four π₯, we get four. And if we differentiate positive
five, it actually just turns to zero. So great! ππ’ ππ₯ is equal to four.

So now weβre gonna find ππ¦
ππ’. And to do that, weβre actually
gonna differentiate ln π’. Weβre going to get one over π’. And we get that because actually
thereβs a general form that tells us that if π¦ is equal to ln π₯, then ππ¦ ππ₯ is
equal to one over π₯.

Okay, great! So now weβve actually
differentiated both parts. We can use the chain rule to put it
together. So weβre gonna get that the
differential of ln four π₯ plus five is equal to four, because that was ππ’ ππ₯
multiplied by one over π’, which is ππ¦ ππ’. So now just substitute in our value
for π’, which gives us four over four π₯ plus five.

Okay, great! So now weβve actually
differentiated our ln four π₯ plus five. Thereβs just one more step to do to
actually find our final ππ¦ ππ₯. So weβre gonna get ππ¦ ππ₯ is
equal to seven multiplied by four over four π₯ plus five. And thatβs, as we said before, was
because we actually differentiated ln four π₯ plus five. So therefore we can say that, given
that π¦ equals ln four π₯ plus five to the power of seven, ππ¦ ππ₯ is equal to 28
over four π₯ plus five.

Itβs worth mentioning at this point
that we couldβve actually used the formula to help us differentiate ln four π₯ plus
five. And that would have been that if it
was in the form π¦ equals ln π π₯, then ππ¦ ππ₯ is equal to the differential of
π π₯ over π π₯. However, I just wanted to
demonstrate how the chain rule is used to actually get to this result. And we can actually double-check it
using this, because if we actually look, our function, so our π π₯, was actually
four π₯ plus five.

So, therefore, if we differentiate
this, we just get four. So that would have been our
numerator. And then on the denominator, it
just wouldβve been the function itself, which was four π₯ plus five. And as we can see in the previous
step, just before the final answer, this was in fact the value that we found when we
differentiated ln four π₯ plus five using the chain rule.