### Video Transcript

Determine the derivative of the
function π of π’ equals π’ squared plus five over π’ squared minus one all to the
power of four.

Letβs begin by considering the
rules that will be useful to us in this question. We can see that we have a quotient,
π’ squared plus five over π’ squared minus one. So, weβre going to need to use the
quotient rule. Iβve written the quotient rule out
here using πs and πs because weβve already got π’s and πs in the question. So, this will enable us to find the
derivative of this expression inside the parentheses.

But we still have that power of
four. Weβll allow π£ to equal the
expression inside the parentheses. Itβs equal to π’ squared plus five
over π’ squared minus one. Then, π will be equal to π£ to the
power of four. And we can apply the chain rule,
dπ by dπ’ is equal to dπ by dπ£ multiplied by dπ£ by dπ’.

dπ by dπ£ can be easily calculated
by applying the power rule. Itβs equal to four π£ cubed. But in order to calculate dπ£ by
dπ’, weβre going to need to apply the quotient rule. Weβll let π equal the numerator of
π£, thatβs π’ squared plus five, and π equals the denominator, thatβs π’ squared
minus one. dπ by dπ’ and dπ by dπ’ can be found by applying the power rule. Theyβre each equal to two π’. Now, we can substitute into the
quotient rule to find dπ£ by dπ’.

We have π, thatβs π’ squared minus
one, multiplied by π prime or dπ by dπ’, thatβs two π’, minus π, thatβs π’
squared plus five, multiplied by π prime or dπ by dπ’, thatβs two π’. Itβs all divided by π squared,
thatβs π’ squared minus one squared. Now, letβs simplify. Expanding the parentheses in the
numerator, we have two π’ cubed minus two π’ minus two π’ cubed minus 10π’. And then the denominator remains π’
squared minus one all squared. The two π’ cubeds will cancel each
other out, leaving negative 12π’ over π’ squared minus one squared.

Weβll now delete some of this
working out for the quotient rule to make some room on the page. We now have then that dπ by dπ£ is
equal to four π£ cubed and dπ£ by dπ’ is equal to negative 12π’ over π’ squared
minus one all squared. So, we can substitute into the
chain rule. dπ by dπ’ is equal to four π£ cubed multiplied by negative 12π’ over π’
squared minus one all squared. Now, remember that dπ by dπ’ must
be in terms of π’ only. So, we need to reverse our
substitution.

π£ is equal to π’ squared plus five
over π’ squared minus one. So, we have four π’ squared plus
five over π’ squared minus one all cubed multiplied by negative 12π’ over π’ squared
minus one all squared. Simplifying gives negative 48π’
multiplied by π’ squared plus five cubed all over π’ squared minus one to the power
of five.

In this question, we saw then that
we needed to apply a combination of the quotient and chain rules in order to find
the derivative of the function π of π’.