### Video Transcript

If π¦ is equal to the fifth root of
five π₯ plus two to the seventh power, find dπ¦ by dπ₯.

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 equal to
the fifth root of the seventh power of a linear function. Thatβs the composition of three
functions. So we could do this by using the
chain rule twice. And this would work. However, thereβs a simpler method
by using our laws of exponents. Weβll use the following two
versions of our laws of exponents.

First, we know the fifth root of π
is just equal to π to the power of one-fifth. Next, we know that π to the power
of π all raised to the power of π is just equal to π to the power of π times
π. Using these, we can show that π¦ is
equal to five π₯ plus two raised to the power of seven over five. But now, we can see that π¦ is the
composition of only two functions. Itβs a linear function raised to
the power of seven over five. So we can differentiate this by
using the chain rule. Weβll set our inner function five
π₯ plus two to be equal to π’. This gives us that π¦ is equal to
π’ to the power of seven over five.

We now recall the following version
of the chain rule. If we have π¦ is a function of π’
and π’ is a function of π₯, then we can find the derivative of π¦ with respect to π₯
by first finding the derivative of π¦ with respect to π’ and multiplying this by the
derivative of π’ with respect to π₯. And this is exactly what we
have. We have that π¦ is a function of
π’. Itβs π’ to the power of seven over
five. And we have that π’ is a function
of π₯. Itβs five π₯ plus two. So we need to find dπ¦ by dπ’ and
dπ’ by dπ₯.

Letβs start by finding dπ¦ by
dπ’. Thatβs the derivative of π’ to the
power of seven over five with respect to π’. And we can do this by using the
power rule for differentiation. We multiply by the exponent of π’,
thatβs seven over five, and reduce this exponent by one. This gives us seven-fifths times π’
to the power of seven-fifths minus one. And seven-fifths minus one is equal
to two-fifths. So we found an expression for dπ¦
by dπ’. Letβs now find an expression for
dπ’ by dπ₯. We have that dπ’ by dπ₯ is the
first derivative of π’ with respect to π₯.

And remember, π’ is equal to five
π₯ plus two. So we want to differentiate five π₯
plus two with respect to π₯. And we can evaluate this by using
the power rule for differentiation. Or we can notice that this is a
linear function. So its slope is equal to the
coefficient of π₯, which is five. So we found expressions for dπ¦ by
dπ’ and dπ’ by dπ₯. We can now use the chain rule. The chain rule tells us dπ¦ by dπ₯
is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And we found that dπ¦ by dπ’ is
equal to seven-fifths π’ to the power of two over five. And dπ’ by dπ₯ is equal to
five. And we can simplify this by
canceling the shared factor of five in the numerator and the denominator, giving us
seven times π’ to the power of two over five.

But remember, weβre finding an
expression for the derivative of π¦ with respect to π₯. So we want our answer to be in
terms of π₯. So weβll use our substitution π’ is
equal to five π₯ plus two. So by using the substitution π’ is
equal to five π₯ plus two, weβve shown that dπ¦ by dπ₯ is equal to seven times five
π₯ plus two to the power of two-fifths. And we could leave our answer like
this. However, remember, our original
expression for π¦ contained the fifth root of a linear function. So we can use both of our laws of
exponents to write our answer in this form. Doing this, we get dπ¦ by dπ₯ is
equal to seven times the fifth root of five π₯ plus two squared.

Therefore, by using the chain rule,
weβve shown if π¦ is equal to the fifth root of five π₯ plus two to the seventh
power, then dπ¦ by dπ₯ is equal to seven times the fifth root of five π₯ plus two
squared.