### Video Transcript

Given that π¦ is equal to the
square root of two π₯ minus five, determine π¦ triple prime.

Weβre given π¦ as a function in π₯,
and weβre asked to determine π¦ triple prime. Thatβs the third derivative of π¦
with respect to π₯. So weβre going to need to
differentiate this expression for π¦ three times. We can see π¦ is given as the
composition of two functions. So one way of doing this would be
to use the chain rule. And this would work. However, because our outer function
is the square root, we can also do this by using the general power rule.

First, by using our laws of
exponents, we can rewrite π¦ as two π₯ minus five all raised to the power of
one-half. And now our outer function is a
power function. So we can differentiate this by
using the general power rule. So weβll start by recalling the
general power rule. This tells us for any
differentiable function π of π₯ and real constant π, the derivative of π of π₯
all raised to the πth power with respect to π₯ is equal to π times π prime of π₯
multiplied by π of π₯ raised to the power of π minus one. In our case, we can see our
exponent π is equal to one-half and our inner function is the linear function two
π₯ minus five.

And now we can see, to use the
general power rule, we need to find an expression for π prime of π₯. Thatβs the derivative of two π₯
minus five with respect to π₯. Of course this is a linear
function, so its derivative will be the coefficient of π₯, which is two. Now, we can substitute π is equal
to one-half, π of π₯ is two π₯ minus five, and π prime of π₯ is two into our
general power rule. This gives us π¦ prime is equal to
one-half times two multiplied by two π₯ minus five all raised to the power of
one-half minus one.

And of course we can simplify
this. First, one-half times two is equal
to one. Next, in our exponent, one-half
minus one is equal to negative one-half. So weβve shown π¦ prime is equal to
two π₯ minus five all raised to the power of negative one-half. Of course we could simplify this
further. However, we need to differentiate
this two more times. And weβre going to do this again by
using the general power rule.

This time, we can see our value of
the exponent π is equal to negative one-half. We can see our inner function π of
π₯ stays the same, so π prime of π₯ will also be the same. So the only difference is, instead
of having π as one-half, weβre going to have π as negative one-half. So by applying the general power
rule with π equal to negative one-half, π of π₯ equal to two π₯ minus five, and π
prime of π₯ equal to two, we get π¦ double prime is equal to negative one-half times
two multiplied by two π₯ minus five all raised to the power of negative one-half
minus one.

And once again we can simplify. Again, we have negative one-half
multiplied by two. This gives us negative one. And in our exponent, we have
negative one-half minus one, which is of course equal to negative three over
two. Therefore, weβve shown π¦ double
prime is equal to negative one times two π₯ minus five all raised to the power of
negative three over two.

And again, we could simplify this
by using our laws of exponents. But we still need to differentiate
this one more time. And we can see we can do this by
using the general power rule. This time, our exponent will be
equal to negative three over two. And we can see our inner function
π of π₯ is the same. So π prime of π₯ will also be the
same. The only difference is our exponent
is negative three over two and weβre multiplying our entire function by negative
one.

So weβll differentiate this one
more time by using the general power rule. We set our exponent π equal to
negative three over two, π of π₯ to be two π₯ minus five, and π prime of π₯ as
once again equal to two. And remember, we have a coefficient
of negative one. This gives us π¦ triple prime is
equal to negative one times negative three over two multiplied by two times two π₯
minus five all raised to the power of negative three over two minus one.

And we can simplify this. First, negative one multiplied by
negative three over two multiplied by two is equal to three. And in our exponent, negative three
over two minus one is equal to negative five over two. Therefore, weβve shown π¦ triple
prime is equal to three times two π₯ minus five all raised to the power of negative
five over two.

And we could leave our answer like
this. However, we donβt need to
differentiate this anymore. So we can simplify this by using
our laws of exponents. First, by using our laws of
exponents, we know π to the power of negative five over two will be equal to one
divided by π to the power of five over two. We could then simplify this even
further by using our laws of exponents. We know π to the power of five
over two is equal to the square root of π to the fifth power. And we can apply this to simplify
π¦ triple prime. We set our value of π equal to two
π₯ minus five. And doing this gives us three
divided by the square root of two π₯ minus five all raised to the fifth power, which
is our final answer.

In this question, we were given
that π¦ was equal to the square root of two π₯ minus five. And we were able to use the general
power rule three times to show that π¦ triple prime was equal to three divided by
the square root of two π₯ minus five all raised to the fifth power.