Video Transcript
Determine the derivative of the
function 𝑔 of 𝑢 equals 𝑢 squared plus five over 𝑢 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,
𝑢 squared plus five over 𝑢 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 𝑞s because we’ve already got 𝑢s 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 𝑢 squared plus five
over 𝑢 squared minus one. Then, 𝑔 will be equal to 𝑣 to the
power of four. And we can apply the chain rule,
d𝑔 by d𝑢 is equal to d𝑔 by d𝑣 multiplied by d𝑣 by d𝑢.
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
d𝑢, we’re going to need to apply the quotient rule. We’ll let 𝑝 equal the numerator of
𝑣, that’s 𝑢 squared plus five, and 𝑞 equals the denominator, that’s 𝑢 squared
minus one. d𝑝 by d𝑢 and d𝑞 by d𝑢 can be found by applying the power rule. They’re each equal to two 𝑢. Now, we can substitute into the
quotient rule to find d𝑣 by d𝑢.
We have 𝑞, that’s 𝑢 squared minus
one, multiplied by 𝑝 prime or d𝑝 by d𝑢, that’s two 𝑢, minus 𝑝, that’s 𝑢
squared plus five, multiplied by 𝑞 prime or d𝑞 by d𝑢, that’s two 𝑢. It’s all divided by 𝑞 squared,
that’s 𝑢 squared minus one squared. Now, let’s simplify. Expanding the parentheses in the
numerator, we have two 𝑢 cubed minus two 𝑢 minus two 𝑢 cubed minus 10𝑢. And then the denominator remains 𝑢
squared minus one all squared. The two 𝑢 cubeds will cancel each
other out, leaving negative 12𝑢 over 𝑢 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 d𝑢 is equal to negative 12𝑢 over 𝑢 squared
minus one all squared. So, we can substitute into the
chain rule. d𝑔 by d𝑢 is equal to four 𝑣 cubed multiplied by negative 12𝑢 over 𝑢
squared minus one all squared. Now, remember that d𝑔 by d𝑢 must
be in terms of 𝑢 only. So, we need to reverse our
substitution.
𝑣 is equal to 𝑢 squared plus five
over 𝑢 squared minus one. So, we have four 𝑢 squared plus
five over 𝑢 squared minus one all cubed multiplied by negative 12𝑢 over 𝑢 squared
minus one all squared. Simplifying gives negative 48𝑢
multiplied by 𝑢 squared plus five cubed all over 𝑢 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 𝑢.
