Chain Rule problem

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

[tex] Find \frac {dy} {dx}[/tex]

[tex]y = \frac {(2x+3)^3} {(4x^2-1)^8}[/tex]

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

[tex] -\frac {2(2x+3)^2(52x^2+96x+3)} {(4x^2-1)^9}[/tex]

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

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  • #2
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First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

[tex]f' (2x+3)^3 = 6(2x+3)^2[/tex]
and
[tex]g' (4x^2-1)^8 = 256x(4x-1)^7[/tex]

Applying the quotient rule I get this

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

I have no clue where they went from there...
 
  • #3
arildno
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g' is wrong, your coefficient in front of the seventh' power parenthesis should be 64, not 256.
 
  • #4
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There should be a "squared" in the (4x - 1) bracket on the right of the top line of the quotient, then it just cancels.
 
  • #5
BobG
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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.

[tex]f' (2x+3)^3 = 6(2x+3)^2[/tex]
and
[tex]g' (4x^2-1)^8 = 256x(4x-1)^7[/tex]

Applying the quotient rule I get this

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

I have no clue where they went from there...
As arildno said, your g' is wrong. Your denominator is [tex](4x^2-1)^{16}[/tex]
You can divide all the terms by a [tex](4x^2-1)^7[/tex]. After that, it's just some factoring and multiplication in the numerator.
 
  • #6
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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! :cry:
 
Last edited:
  • #7
TD
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Did you get there now or still need help?
 
  • #8
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Wait, I already did that, never mind...
 
  • #9
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Ok am I on the right track here?

[tex]\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}[/tex]
 
  • #10
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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
 
  • #11
TD
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Yes, seems to be correct so far :smile:
 
  • #12
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Was that to my last post? Because I somehow doubt it :(
 
  • #13
TD
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It was in referrence to
Aresius said:
Ok am I on the right track here?

[tex]\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}[/tex]
So you're doing ok so far.
 
  • #14
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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?
 
  • #15
TD
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You have a bit too many 8's then ...

[tex]\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 }}[/tex]

One comes from the power (x^8 => 8x^7) and the other one from (4x²)' = 8x, which gives 8*8x = 64x.
 
  • #16
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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.
 
  • #17
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[tex]8(4x^2-1)^7(8x)(8)[/tex] 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...
 
  • #18
TD
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We're finished at that point, no more 8's to multiply...

[tex]\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 [/tex]

Ah no, you have an 8 extra. The derivative of 4x² is just 8x; not 8x*8 since (4x²)' = 4*(x²)' = 4*2x = 8x.
 
  • #19
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Am I right so far in simplifying (using pascal's triangle and binomial theorem)

[tex]\frac {6(4x^2-1)(2x+3)^2 - 64x(8x^3+36x^2+54x+27)} {4x^2-1)^9}[/tex]
 
  • #20
TD
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Yes, that is correct. You also could've factored out [itex](2x+3)^2[/itex] first.
 
  • #21
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The answer to the problem still has the (2x+3)^2 so I left it as it is, am I wrong?
 
  • #22
TD
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It's not wrong but it'll be easier to factor it out. As you say, the factor is still there in the answer.

[tex]\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 }}[/tex]

Now work out everything in those last parenthesis.
 
  • #23
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Got it!

Thanks!
 
  • #24
TD
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Great :smile:
 
  • #25
BobG
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It's sometimes easier to use substitution to organize yourself.

I know you don't think the derivative of [tex]4x^2-1[/tex] 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:

[tex]4x^2-1 = u[/tex] then you have two problems:

You need to find the derivative of [tex]u^8[/tex]

You need to find the derivative of [tex]4x^2-1[/tex]
 

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