# 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.

#### 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...

#### arildno

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.

#### BobG

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:

#### TD

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

#### TD

Homework Helper
Yes, seems to be correct so far

#### Aresius

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

#### TD

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?

#### TD

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...

#### TD

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)

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

#### TD

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

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