### Video Transcript

Find dπ¦ by dπ₯ given that π¦ is
equal to nine π₯ divided by the natural logarithm of nine π₯.

The question wants us to find dπ¦
by dπ₯. Thatβs the first derivative of π¦
with respect to π₯. And we can see that π¦ is the
quotient of two functions. Itβs the quotient of nine π₯ and
the natural logarithm of nine π₯. So weβll find this derivative by
using the quotient rule. We recall the quotient rule tells
us if π¦ is the quotient of two functions π’ over π£, then dπ¦ by dπ₯ is equal to π£
times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all divided by π£ squared.

So to use the quotient rule, weβll
start by setting π’ of π₯ to be the function in our numerator, thatβs nine π₯, and
π£ of π₯ to be the function in our denominator, thatβs the natural logarithm of nine
π₯. And to apply the quotient rule,
weβre going to need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with finding dπ’ by
dπ₯. Thatβs the derivative of nine π₯
with respect to π₯. And nine π₯ is just a linear
function. So its derivative is the
coefficient of π₯, which, in this case, is nine.

Letβs now find an expression for
dπ£ by dπ₯. Thatβs the derivative of the
natural logarithm of nine π₯ with respect to π₯. And we can do this by using one of
our standard derivative results for logarithmic functions. For any positive constant π, the
derivative of the natural logarithm of ππ₯ with respect to π₯ is equal to one
divided by π₯. So in our case, the derivative of
the natural logarithm of nine π₯ with respect to π₯ is equal to one divided by
π₯. So weβre now ready to find the dπ¦
by dπ₯ by using the quotient rule.

The quotient rule tells us dπ¦ by
dπ₯ will be equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ divided by π£
squared. Substituting in our expressions for
π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get dπ¦ by dπ₯ is equal to the natural
logarithm of nine π₯ multiplied by nine minus nine π₯ times one over π₯ all divided
by the natural logarithm of nine π₯ squared. And we can simplify this
expression. First, weβll cancel π₯ multiplied
by one over π₯.

Next, we want to take out a factor
of nine in our numerator. And this gives us nine times the
natural logarithm of nine π₯ minus one all divided by the natural logarithm of nine
π₯ squared. And this is our final answer. Therefore, weβve shown if π¦ is
equal to nine π₯ divided by the natural logarithm of nine π₯, then dπ¦ by dπ₯ is
equal to nine times the natural logarithm of nine π₯ minus one all divided by the
natural logarithm of nine π₯ squared.