### Video Transcript

Find dπ¦ by dπ₯ if six π¦ is equal
to seven π₯ raised to the power three over five π₯.

Weβre asked to find the derivative
with respect to π₯ of the function six π¦ is equal to seven π₯ raised to the power
of three over five π₯. And before we begin, there are a
couple of things we need to note about our equation here. The first is that on the left-hand
side, we have six π¦ as opposed to π¦. Since π¦ is not by itself on the
left, we call this an implicit function or equation. The second thing to notice that our
exponent on the right-hand side is not a constant. In fact, our exponent involves a
variable π₯. And because of this, we canβt use
any of our usual rules for differentiation. What we can do to find dπ¦ by dπ₯
is to use logarithmic differentiation.

The first thing we can do is to
isolate π¦ on the left-hand side by dividing through by six. We can then cancel the sixes on the
left so that we have π¦ is equal to seven over six π₯ raised to the power three over
five π₯. We now apply logarithmic
differentiation. And our first step is to apply the
natural logarithm to both sides. And remember that the natural
logarithm is the logarithm to the base π, where, to five decimal places, π which
is Eulerβs number is 2.71828 so that we have the natural logarithm of π¦ is the
natural logarithm of π of π₯. In our case, thatβs the natural
logarithm of seven over six times π₯ raised to the power three over five π₯.

We should note that for the
logarithmic differentiation to be valid, we need to specify here that π¦ is greater
than zero. Thatβs because the logarithm of
zero is undefined and the function doesnβt exist for negative values. If we want to include negative
values, we must include absolute value lines around π¦ and around π of π₯. But in our case, weβll simply
assume that π¦ is greater than zero. Now, our function actually looks
more complicated than it did in the first place. But this is where the use of the
logarithms comes in. We use the laws of logarithms to
expand our right-hand side. The first thing we can do is use
the product rule for logs, which says that the log of ππ is equal to the log of π
plus the log of π, so that on our right-hand side, we can split this up into the
log of seven over six plus the natural log of π₯ raised to the power three over five
π₯.

And we can expand further our
second term using the power rule for logs. Thatβs the log of π raised to the
power π is π times the log of π. Our right-hand side is then the
natural logarithm of seven over six plus three over five π₯ times the natural
logarithm of π₯. Our third step in logarithmic
differentiation is to differentiate both sides with respect to π₯. On our left-hand side, weβre
differentiating a function of a function because π¦ is actually a function of
π₯. And remember that if we have a
function π which is a function of π¦ which is a function of π₯, by the chain rule,
we have dπ by dπ₯ is equal to dπ by dπ¦ times dπ¦ by dπ₯. And in our case, π is the natural
logarithm of π¦.

And now, we can use the result that
the derivative of the natural logarithm of π’ with respect to π’ is one over π’ for
π’ greater than zero so that on our left-hand side, we have one over π¦ times dπ¦ by
dπ₯, where one over π¦ is our dπ by dπ¦. And now, on the right-hand side, we
know that the logarithm of seven over six is the logarithm of a constant is a
constant, so the derivative of this is equal to zero. So now, we need to find the
derivative of three over five π₯ times the logarithm of π₯. And for this, we can use the
product rule for differentiation, where, in our case, π is three over five π₯ and
π is the natural logarithm of π₯.

Noting that three over five π₯ is
actually three over five times π₯ raised to the power negative one and this is a
function of the form π times π₯ raised to the πth power, then we can use the power
rule for derivatives, which says that d by dπ₯ of π times π₯ raised to the πth
power is π times π times π₯ raised to the π minus oneth power. That is, we multiply by the
exponent and subtract one from the exponent. d by dπ₯ is three over five π₯ raised
to the power negative one is negative three over five π₯ raised to the power
negative two.

And recalling that the derivative
of the natural logarithm of π₯ is one over π₯, so that in our product rule π prime
is negative three over five π₯ raised to the power negative two and π prime is one
over π₯, the derivative on our right-hand side is negative three over five π₯
squared, which is π prime, times the natural logarithm of π₯, which is π, plus
three over five π₯, which is π, times one over π₯, which is π prime. Rearranging this, we have three
over five π₯ squared minus three over five π₯ squared times the natural logarithm of
π₯. And since we have a common factor
of three over five π₯ squared, we can take this outside.

Now, remember, weβre trying to find
dπ¦ by dπ₯. This means we need to isolate dπ¦
by dπ₯ on the left-hand side. We can do this by multiplying
through by π¦. And this is our fourth step in our
logarithmic differentiation. We can cancel the π¦βs on the
left-hand side. And on the right-hand side, we have
three π¦ over five π₯ squared times one minus the logarithm of π₯. Remember, though, that π¦ is equal
to seven over six times π₯ raised to the power three over five π₯. And substituting this in, we can
cancel a three so that our constant is actually seven over 10. So now, if we look at our powers of
π₯, this is actually π₯ raised to the power three over five π₯ minus two. And we have dπ¦ by dπ₯ is equal to
seven over 10 times π₯ raised to the power three over five π₯ minus two times one
minus the natural logarithm of π₯.