Video: Differentiating Rational Functions Using the Quotient and Chain Rules

Determine the derivative of the function 𝑔(𝑒) = ((𝑒² + 5)/(𝑒² βˆ’ 1))⁴.

03:31

Video Transcript

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

Let’s begin by considering the rules that will be useful to us in this question. We can see that we have a quotient, 𝑒 squared plus five over 𝑒 squared minus one. So, we’re going to need to use the quotient rule. I’ve written the quotient rule out here using 𝑝s and π‘žs because we’ve already got 𝑒s and 𝑔s 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. We’ll allow 𝑣 to equal the expression inside the parentheses. It’s equal to 𝑒 squared plus five over 𝑒 squared minus one. Then, 𝑔 will be equal to 𝑣 to the power of four. And we can apply the chain rule, d𝑔 by d𝑒 is equal to d𝑔 by d𝑣 multiplied by d𝑣 by d𝑒.

d𝑔 by d𝑣 can be easily calculated by applying the power rule. It’s equal to four 𝑣 cubed. But in order to calculate d𝑣 by d𝑒, we’re going to need to apply the quotient rule. We’ll let 𝑝 equal the numerator of 𝑣, that’s 𝑒 squared plus five, and π‘ž equals the denominator, that’s 𝑒 squared minus one. d𝑝 by d𝑒 and dπ‘ž by d𝑒 can be found by applying the power rule. They’re each equal to two 𝑒. Now, we can substitute into the quotient rule to find d𝑣 by d𝑒.

We have π‘ž, that’s 𝑒 squared minus one, multiplied by 𝑝 prime or d𝑝 by d𝑒, that’s two 𝑒, minus 𝑝, that’s 𝑒 squared plus five, multiplied by π‘ž prime or dπ‘ž by d𝑒, that’s two 𝑒. It’s all divided by π‘ž squared, that’s 𝑒 squared minus one squared. Now, let’s simplify. Expanding the parentheses in the numerator, we have two 𝑒 cubed minus two 𝑒 minus two 𝑒 cubed minus 10𝑒. And then the denominator remains 𝑒 squared minus one all squared. The two 𝑒 cubeds will cancel each other out, leaving negative 12𝑒 over 𝑒 squared minus one squared.

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

𝑣 is equal to 𝑒 squared plus five over 𝑒 squared minus one. So, we have four 𝑒 squared plus five over 𝑒 squared minus one all cubed multiplied by negative 12𝑒 over 𝑒 squared minus one all squared. Simplifying gives negative 48𝑒 multiplied by 𝑒 squared plus five cubed all over 𝑒 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 𝑔 of 𝑒.

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