# Thread: Why are they using the chain rule on this question?

1. ## Why are they using the chain rule on this question?

I'm stuck on a question from my chapter review homework. It appears the textbook has used the Chain Rule, but we were never taught how to use the Chain Rule in this context, so I don't know how to apply it.

How would I work this question out?

Find the derivative.
$y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}$

I tried to chain the top and then the bottom separately- obviously that can't be right.

2. Originally Posted by NAPA55
I'm stuck on a question from my chapter review homework. It appears the textbook has used the Chain Rule, but we were never taught how to use the Chain Rule in this context, so I don't know how to apply it.

How would I work this question out?

Find the derivative.
$y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}$

I tried to chain the top and then the bottom separately- obviously that can't be right.
we use the chain rule to find the derivative of composite functions. clearly, $(x - 4)^5$ and $(2x^3 - 1)^2$ are composite functions. you use the chain rule when differentiating them. otherwise, you would use the quotient and product rules to differentiated the whole function

3. This is what I tried to do:

I tried to chain the top to find the derivative of f(x), which is the top function.

I tried to chain the bottom to find the derivative of g(x), which is the bottom function.

I got this, and I tried to use the quotient rule after I used the chain rule.
$f(x) = (x+2)(x-4)^5$ $f '(x) = (x-4)^5 + 5(x+2)(x-4)^4$

$g(x) = (2x^3-1)^2$ $g '(x) = 2(2x^3-1)(6x) = 12x(2x^3-1)$

But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
$y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}$ $[\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]$

4. ## logarithmic differentiation

It would be a combination of rules chain quotent ect.

Why not try logarithmic differentiation
$
y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}
$

$
ln(y) = ln \left(\frac{(x+2)(x-4)^5}{(2x^3-1)^2} \right)=ln(x+2)+5ln(x-4)-2ln(2x^3-1)
$

taking the derivative

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{(x+2)}+\frac{5}{ (x-4)}-\frac{12x^2}{(2x^3-1)}$

Solving for $\frac{dy}{dx}$ gives

$\frac{dy}{dx}=y \left( \frac{1}{(x+2)}+\frac{5}{(x-4)}-\frac{12x^2}{(2x^3-1)}\right)=$

$\frac{dy}{dx}= \frac{(x+2)(x-4)^5}{(2x^3-1)^2} \cdot \left( \frac{1}{(x+2)}+\frac{5}{(x-4)}-\frac{12x^2}{(2x^3-1)}\right)$

5. If logarithmic differentiation is needed to do that question, we skipped that topic in this unit, and as far as I know, we're not responsible for it.

If this is a logarithmic differentiation type question that was on the review, I guess I shouldn't have attempted it yet. This always happens to me! I try a question, get frustrated because I can't do it, and then later learn it wasn't something we were taught yet! LOL

6. Originally Posted by NAPA55
This is what I tried to do:

I tried to chain the top to find the derivative of f(x), which is the top function.

I tried to chain the bottom to find the derivative of g(x), which is the bottom function.

I got this, and I tried to use the quotient rule after I used the chain rule.
$f(x) = (x+2)(x-4)^5$ $f '(x) = (x-4)^5 + 5(x+2)(x-4)^4$

$g(x) = (2x^3-1)^2$ $g '(x) = 2(2x^3-1)(6x) = 12x(2x^3-1)$

But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
$y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}$ $[\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]$
your derivative for the second is not correct, you should have 6x^2 not 6x. maybe that's why you have been having problems.

Originally Posted by NAPA55
If logarithmic differentiation is needed to do that question, we skipped that topic in this unit, and as far as I know, we're not responsible for it.

If this is a logarithmic differentiation type question that was on the review, I guess I shouldn't have attempted it yet. This always happens to me! I try a question, get frustrated because I can't do it, and then later learn it wasn't something we were taught yet! LOL
logarithmic differentiation is not REQUIRED, it is just another way to do the problem. this problem can be done with only the chain, quotient and product rules.

7. That was a typo on my part. It should be $12x^2(2x^3-1).$ Should it work if I keep going from what I had?

8. Originally Posted by NAPA55

But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
$y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}$ $[\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]$
Because of the form of the book answer I would guess "they" (The book )used logarithmic differentiation

If you have time to learn it, it is a great tool for taking derivatives.

Good luck.

9. Originally Posted by NAPA55
That was a typo on my part. It should be $12x^2(2x^3-1).$ Should it work if I keep going from what I had?
yes, keep going.

TheEmptySet has a good point. it would seem that the book did use logarithmic differentiation. it can be done without it though, so continue. i also agree with TheEmptySet in that, if you can learn logarithmic differentiation by yourself, you should. you are familiar with the laws of logarithms right?