Explaining the chain rule when using quotient rule.

Question

I have a problem in my WebWorks hw , modeled after Rogawski ET 3e section 3.7, exercise 35. The problem reads, find the derivative of $(\frac{x+81}{x-81})^{20}$. I know the answer is $-3240 \frac{(x+81)^{19}}{(x-81)^{21}}$, however I don't know how to get that answer. Can someone explain the steps to achieving the answer?

Answer 1

We need to find the derivative of $$f(x)=\left(\frac{x+81}{x-81}\right)^{20}$$ We must use the chain rule here. Let $u=\dfrac{x+81}{x-81}$ so the function now becomes $$f(u)=u^{20}$$ The derivative of this function is $f'(u)\cdot u'$(definition of chain rule). We know $f'(u)$ to be $20u^{19}$. Now we must multiply by $u'$. \begin{align*} u'&=\left(\frac{x+81}{x-81}\right)'\\ &= \frac{x-81-(x+81)}{(x-81)^2} \\ &= -\frac{162}{(x-81)^2} \end{align*} So the derivative is \begin{align*} f'(x)&=20u^{19}\cdot -\frac{162}{(x-81)^2}\\ &=-20\left(\frac{x+81}{x-81}\right)^{19}\cdot \frac{162}{(x-81)^2}\\ &=-20\frac{(x+81)^{19}}{(x-81)^{19}}\cdot \frac{162}{(x-81)^2}\\ &= -\frac{3240(x+81)^{19}}{(x-81)^{21}} \end{align*}

Answer 2

As you expect, you have to apply both the chain rule and the quotient rule. Let's start by using the chain rule to represent $\frac{x+81}{x-81}$ as a function $u(x)$. Then what you are trying to find is, by the chain rule,

$$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $$

Where $y = (\frac{x+81}{x-81})^{20} = u^{20}$.

So, as @Peter showed,

$$ \frac{dy}{du} = 20u^{19} = 20(\frac{x+81}{x-81})^{19} $$

And all you need to do to find $\frac{dy}{dx}$ is to multiply that by $\frac{du}{dx}$, which you can find yourself using the quotient rule,

$$ \frac{du}{dx} = \frac{d}{dx} (\frac{x+81}{x-81}) $$

Good luck!