### Video Transcript

Given that π¦ is equal to three
π₯ squared minus five over two π₯ squared plus seven, determine the second
derivative of π¦ with respect to π₯.

Here we have a quotient. Itβs the result of dividing one
function by another function. We can, therefore, use the
quotient rule to help us find the first derivative. This says that for two
differentiable functions π’ and π£, the derivative of π’ over π£ with respect to
π₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£
squared.

Since π’ is the numerator,
weβre going to let π’ be equal to three π₯ squared minus five. And we can see that π£ must be
equal to two π₯ squared plus seven. To be able to use the quotient
rule, weβre going to differentiate each of these with respect to π₯. When we differentiate π’ with
respect to π₯, we get six π₯. And dπ£ by dπ₯ is equal to four
π₯. π£ times dπ’ by dπ₯ is two π₯
squared plus seven times six π₯. We then subtract the product of
π’ and dπ£ by dπ₯. Thatβs three π₯ squared minus
five times four π₯. And, of course, thatβs all over
π£ squared. Thatβs two π₯ squared plus
seven squared.

Distributing the parentheses on
the top of our fraction, and we get 12π₯ cubed plus 42π₯ minus 12π₯ cubed plus
20π₯. And then, weβll keep the
denominator as is. And we see that the derivative
of π¦ with respect π₯ is 62π₯ over two π₯ squared plus seven squared.

To find the second derivative,
weβre going to need to differentiate this again. Letβs clear some space. Once again, weβre looking to
differentiate a quotient, so weβre going to use the quotient rule. This time, we let π’ be equal
to 62π₯ and π£ be equal to two π₯ squared plus seven squared. dπ’ by dπ₯ is
fairly straightforward. Itβs 62. But what about dπ£ by dπ₯?

Well, we can use a special case
of the chain rule called the general power rule. This says that if π is some
function of π₯ and π is some constant rational not equal to zero, then the
derivative of π to the power of π with respect to π₯ is π times π to the
power of π minus one times dπ by dπ₯. This means that dπ£ by dπ₯ is
two times this function in π₯, which is two π₯ squared plus seven to the power
of one, times the derivative of two π₯ squared plus seven with respect π₯, which
is four π₯.

And when we simplify this, we
see that dπ£ by dπ₯ is eight π₯ times two π₯ squared plus seven. This time, π£ times dπ’ by dπ₯
is two π₯ squared plus seven squared times 62 minus π’ times dπ£ by dπ₯ all over
π£ squared. The key here is to spot that
the denominator of our fraction becomes two π₯ squared plus seven to the power
of four. And this means we can divide
through by a common factor of two π₯ squared plus seven. And that leaves us with two π₯
squared plus seven cubed on the denominator and 62 times two π₯ squared plus
seven minus 62π₯ times eight π₯ on the top.

We can take out a factor of 62
on the numerator. And then, when we subtract
eight π₯ times π₯ from two π₯ squared, we get negative six π₯ squared. And the second derivative of π¦
with respect to π₯ is 62 times seven minus six squared over two π₯ squared plus
seven cubed.