### Video Transcript

Find dπ¦ by dπ₯ if π¦ is equal to
six times π₯ to the ninth power minus five all multiplied by the natural logarithm
of π₯.

Weβre given an equation for π¦ in
terms of π₯. And we need to use this to find an
expression for dπ¦ by dπ₯. Thatβs the derivative of π¦ with
respect to π₯. To do this, we need to notice that
π¦ is the product of two functions. Itβs six π₯ to the ninth power
minus five multiplied by the natural logarithm of π₯. And we know how to differentiate
both of these expressions. So we can evaluate this derivative
by using the product rule.

So letβs start by recalling the
product rule. We know if π¦ is the product of two
differentiable functions π of π₯ times π of π₯, then the product rule tells us dπ¦
by dπ₯ would be equal to π prime of π₯ times π of π₯ plus π prime of π₯ times π
of π₯. And we can see this is exactly what
we have in this question. Weβll set π of π₯ to be six π₯ to
the ninth power minus five and π of π₯ to be the natural logarithm of π₯.

So we can find dπ¦ by dπ₯ by using
the product rule. However, we see to use the product
rule, weβre going to need to find expressions for π prime of π₯ and π prime of
π₯. Letβs start by finding an
expression for π prime of π₯. Thatβs the derivative of π of π₯
with respect to π₯, which is the derivative of six π₯ to the ninth power minus five
with respect to π₯. And we can see this is the
derivative of a polynomial. So we can do this term by term by
using the power rule for differentiation.

We want to multiply by our exponent
of π₯ and reduce this exponent by one. In our first term, this gives us
nine times six π₯ to the eighth power. And in our second term, we know the
derivative of the constant negative five with respect to π₯ will be equal to
zero. And of course, we can simplify this
expression for π prime of π₯. We get nine times six is equal to
54. So weβve shown π prime of π₯ is
equal to 54π₯ to the eighth power.

Letβs now find an expression for π
prime of π₯. Once again, thatβs the derivative
of π of π₯ with respect to π₯, which is the derivative of the natural logarithm of
π₯ with respect to π₯. And to evaluate this derivative, we
need to recall one of our standard derivative results. The derivative of the natural
logarithm of π₯ with respect to π₯ is equal to the reciprocal function one over
π₯. So by using this, weβve shown that
π prime of π₯ is equal to one over π₯.

Weβre now ready to find an
expression for dπ¦ by dπ₯ by using our formula from the product rule. We just need to substitute our
expressions for π of π₯, π of π₯, π prime of π₯, and π prime of π₯ into our
formula from the product rule. Doing this, we get dπ¦ by dπ₯ is
equal to 54π₯ to the eighth power times the natural logarithm of π₯ plus one over π₯
multiplied by six π₯ to the ninth power minus five.

And we could leave our answer like
this. However, we could also
simplify. Weβll do this by distributing one
over π₯ over the parentheses in our second term. This gives us 54π₯ to the eighth
power multiplied by the natural logarithm of π₯ plus six π₯ to the ninth power
divided by π₯ minus five over π₯. And of course we can simplify
this. We have six π₯ to the ninth power
divided by π₯. Using our laws of exponents, this
is just equal to six π₯ to the eighth power.

And thereβs one more piece of
simplification we can do. We can take out the shared factor
of six π₯ to the eighth power in our first and second term. And doing this gives us six π₯ to
the eighth power multiplied by nine times the natural logarithm of π₯ plus one minus
five over π₯. And this gives us our final
answer.

Therefore, we were able to show if
π¦ is equal to six π₯ to the ninth power minus five all multiplied by the natural
logarithm of π₯, then by using the product rule, we can show that dπ¦ by dπ₯ must be
equal to six π₯ to the eighth power multiplied by nine times the natural logarithm
of π₯ plus one minus five divided by π₯.