# Chain Rule problem

Aresius
This is a problem that has stumped my entire class of Calc 1 students and two Calc 2 students.

$$Find \frac {dy} {dx}$$

$$y = \frac {(2x+3)^3} {(4x^2-1)^8}$$

I know that the answer is (from the textbook, but I don't know how it got there)

$$-\frac {2(2x+3)^2(52x^2+96x+3)} {(4x^2-1)^9}$$

I'll attempt to show my work in the next post.

## Answers and Replies

Aresius
First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

$$f' (2x+3)^3 = 6(2x+3)^2$$
and
$$g' (4x^2-1)^8 = 256x(4x-1)^7$$

Applying the quotient rule I get this

$$\frac {6(4x^2-1)^8(2x+3)^2 - 256x(4x-1)^7(2x+3)^3} {[(4x^2-1)^8]^2}$$

I have no clue where they went from there...

Homework Helper
Gold Member
Dearly Missed
g' is wrong, your coefficient in front of the seventh' power parenthesis should be 64, not 256.

Mulder
There should be a "squared" in the (4x - 1) bracket on the right of the top line of the quotient, then it just cancels.

Homework Helper
Aresius said:
First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

$$f' (2x+3)^3 = 6(2x+3)^2$$
and
$$g' (4x^2-1)^8 = 256x(4x-1)^7$$

Applying the quotient rule I get this

$$\frac {6(4x^2-1)^8(2x+3)^2 - 256x(4x-1)^7(2x+3)^3} {[(4x^2-1)^8]^2}$$

I have no clue where they went from there...
As arildno said, your g' is wrong. Your denominator is $$(4x^2-1)^{16}$$
You can divide all the terms by a $$(4x^2-1)^7$$. After that, it's just some factoring and multiplication in the numerator.

Aresius
I thought that because of the chain rule you have to take the derivative of the derivative of 4x^2 - 1 and multiply it by the others in sequence.

God I hate the chain rule...

I think I messed up g' more than I thought... You cannot touch the damn original function!

Last edited:
Homework Helper
Did you get there now or still need help?

Aresius
Wait, I already did that, never mind...

Aresius
Ok am I on the right track here?

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}$$

Aresius
You must take the derivative of 8x as per the chain rule and then multiply it into that term.

Am I right?

In that case I would have a 512 in front of that x, bleh

Homework Helper
Yes, seems to be correct so far

Aresius
Was that to my last post? Because I somehow doubt it :(

Homework Helper
It was in referrence to
Aresius said:
Ok am I on the right track here?

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}$$
So you're doing ok so far.

Aresius
But that I don't get, as per the chain rule you must first get 8 from g', then 8x from what was in the brackets, then another 8 from 8x. Multiply them together and you get 512x, why is it still 64x?

Homework Helper
You have a bit too many 8's then ...

$$\left( {\frac{{\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^8 }}} \right)^\prime = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right)^8 - 64x\left( {4x^2 - 1} \right)^7 \left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^{16} }} = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right) - 64x\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^9 }}$$

One comes from the power (x^8 => 8x^7) and the other one from (4x²)' = 8x, which gives 8*8x = 64x.

Aresius
Yes, and then you must take the derivative of 8x and multiply that on because of the chain rule, right?

I thought that as long as you can take a rational derivative with the chain rule, you should.

Aresius
$$8(4x^2-1)^7(8x)(8)$$ is how we do chain rule in class.

Perhaps I'm horribly mistaken and I've forgotten part of the rules of the chain rule...

Homework Helper
We're finished at that point, no more 8's to multiply...

$$\left( {\left( {4x^2 - 1} \right)^8 } \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot \left( {4x^2 - 1} \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot 8x = 64x\left( {4x^2 - 1} \right)^7$$

Ah no, you have an 8 extra. The derivative of 4x² is just 8x; not 8x*8 since (4x²)' = 4*(x²)' = 4*2x = 8x.

Aresius
Am I right so far in simplifying (using pascal's triangle and binomial theorem)

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(8x^3+36x^2+54x+27)} {4x^2-1)^9}$$

Homework Helper
Yes, that is correct. You also could've factored out $(2x+3)^2$ first.

Aresius
The answer to the problem still has the (2x+3)^2 so I left it as it is, am I wrong?

Homework Helper
It's not wrong but it'll be easier to factor it out. As you say, the factor is still there in the answer.

$$\frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right) - 64x\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^9 }} = \frac{{2\left( {2x + 3} \right)^2 \left( {3\left( {4x^2 - 1} \right) - 32x\left( {2x + 3} \right)} \right)}} {{\left( {4x^2 - 1} \right)^9 }}$$

Now work out everything in those last parenthesis.

Aresius
Got it!

Thanks!

Homework Helper
Great

Homework Helper
It's sometimes easier to use substitution to organize yourself.

I know you don't think the derivative of $$4x^2-1$$ is 8*8x. Your problem was the same a lot of people have with the chain rule - once they get going, they don't know when to stop.

If you had broken it out as:

$$4x^2-1 = u$$ then you have two problems:

You need to find the derivative of $$u^8$$

You need to find the derivative of $$4x^2-1$$

Aresius
I don't know why I thought this, but I thought as long as you had a term that could be counted as a function inside the original function, you should take it's derivative and multiply in series as per chain rule.

I think I see what you mean though, you only take derivative of what has not been derived already. The original function never changes.