# Video: Differentiating Rational Functions Using the Quotient and Chain Rules

Determine the derivative of the function π(π’) = ((π’Β² + 5)/(π’Β² β 1))β΄.

03:31

### 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.

Letβs 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, weβre going to need to use the quotient rule. Iβve written the quotient rule out here using πs and πs because weβve already got π’s and πs 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. Weβll allow π£ to equal the expression inside the parentheses. Itβs 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. Itβs equal to four π£ cubed. But in order to calculate dπ£ by dπ’, weβre going to need to apply the quotient rule. Weβll let π equal the numerator of π£, thatβs π’ squared plus five, and π equals the denominator, thatβs π’ squared minus one. dπ by dπ’ and dπ by dπ’ can be found by applying the power rule. Theyβre each equal to two π’. Now, we can substitute into the quotient rule to find dπ£ by dπ’.

We have π, thatβs π’ squared minus one, multiplied by π prime or dπ by dπ’, thatβs two π’, minus π, thatβs π’ squared plus five, multiplied by π prime or dπ by dπ’, thatβs two π’. Itβs all divided by π squared, thatβs π’ squared minus one squared. Now, letβs 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.

Weβll 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’.