### Video Transcript

Using logarithmic differentiation,
determine the derivative of π¦ is equal to the square root of π₯ plus one over two
π₯ to the power four minus two.

Weβre given what looks to be a
fairly complicated function π¦, which is the square root of a quotient function. And we can write this as π₯ plus
one over two π₯ to the of power four minus two all raised to the power of a half
since an exponent of a half is the same thing as a square root. Now we could treat this as a
function of a function and use the chain rule and the quotient rule to
differentiate. However, weβre specifically asked
to use logarithmic differentiation.

But how does this work? Well, letβs see. Suppose we have a function π¦ is
equal to π of π₯. Then the first thing we do is apply
the natural logarithm to both sides. We then have the natural logarithm
of π¦ is equal to the natural logarithm of π of π₯, recalling that the natural
logarithm means the log to the base π, where π is Eulerβs number, which is
approximately 2.71828 and so on. And for this to be valid, we need
to specify that π¦ is greater than zero, since the log of zero is undefined and the
log doesnβt exist for negative values.

If we did want to include negative
values, we use the absolute values of π¦ and π of π₯ and then specify that π¦ is
not equal to zero. Our second step in logarithmic
differentiation is to use the laws of logarithms to expand the natural log of π of
π₯. This should give us something that
we can easily differentiate. And so we differentiate on both
sides with respect to π₯. And once weβve done this, we can
solve for π¦ prime. Thatβs dπ¦ by dπ₯. So letβs begin working through our
steps with our function π¦ by taking the natural logarithms on both sides.

And in our case, we can specify
that π¦ is the positive square root. Next, we use the laws of logarithms
to expand our right-hand side. And the first law we can use since
we have an exponent of a half on our right-hand side is the power rule for
logarithms. This says that log to the base π
of π raised to the power π is π times log to the base π of π. That is, we bring our exponent to
the front and multiply by it. In our case then, this means we
bring our half down to the front.

Now the argument of our natural
logarithm on the right-hand side is a quotient. And we can simplify this by using
the quotient rule for logarithms that says log to the base π of π over π is the
log to the base π of π minus log to the base π of π. In our case, π₯ plus one
corresponds to π and two π₯ to the power of four minus two corresponds to π. And so we have one over two times
the natural logarithm of π₯ plus one minus the natural logarithm of two π₯ to the
power of four minus two. So now we have something manageable
on our right-hand side that we know how to differentiate quite easily.

Now we want the derivative with
respect to π₯. And on our left-hand side, weβre
differentiating a function of a function because, of course, π¦ is a function of
π₯. And on our right-hand side, the
derivative of a sum is the sum of the derivatives. So we can split this into two
derivatives and take the constant a half out front. And on both sides, weβre going to
use the known result that d by dπ₯ of the natural logarithm of π’, where π’ is a
differentiable function of π₯, is one over π’ times dπ’ by dπ₯. And thatβs for π’ greater than
zero.

On our left-hand side, this gives
us one over π¦ dπ¦ by dπ₯. And on our right-hand side, in our
first term, if we have π’ is equal to π₯ plus one and dπ’ by dπ₯ is equal to one,
then d by dπ₯ of the natural logarithm of π₯ plus one is one times one over π₯ plus
one. Similarly, in our second term, if
we let π’ equal to two π₯ to the power of four minus two, then dπ’ by dπ₯ is eight
times π₯ to the power three. And so the derivative with respect
to π₯ of the natural logarithm of two π₯ raised to power of four minus two is eight
π₯ cubed times one over two π₯ to the power of four minus two.

And writing this out a little more
neatly, we have one over π¦ dπ¦ by dπ₯ on our left-hand side is equal to one over
two times one over π₯ plus one minus eight π₯ cubed over two π₯ to the power of four
minus two. In the second term, on our
right-hand side, in the denominator, we have a common factor of two, which we can
cancel with the factor two in the numerator. And we can also bring our half
inside our parentheses. And we can note again that in the
second term in our right-hand side, we have a common factor of two in the numerator
and denominator. And so we have one over π¦ dπ¦ by
dπ₯ is one over two π₯ plus two minus two π₯ to the power of three over π₯ to the
power of four minus one.

Now making some room, we can note
that we have a factor of one over π¦ on the left-hand side, so weβre not quite
finished. To eliminate this, we can multiply
both sides by π¦. And the π¦βs on the left-hand side
will then cancel. And if we then substitute back in
our function π¦, we have dπ¦ by dπ₯ is as shown. And if we rewrite this as a square
root as itβs given in the question, we have dπ¦ by dπ₯, so thatβs the derivative of
π¦ with respect to π₯, is the square root of π₯ plus one over two π₯ to the power of
four minus two times one over two π₯ plus two minus two π₯ cubed over π₯ to the
power of four minus one.