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Chain rule

  1. Mar 16, 2014 #1
    Differentiate the following by rule y=(2x2+4x)5

    Is the chain rule the right rule to use?


    Let U=2x2+4x


    y=(u)5 → dy/du=5(u)4



    dy/dx= 30(2x2+4x)44x

    dy/dx= 30(2x216x)4

    I'm wondering if I am on the right track?
  2. jcsd
  3. Mar 16, 2014 #2


    Staff: Mentor

    Anthonyk, your questions should be posted in the Homework & Coursework sections (Calculus & Beyond) - not in the technical math sections.
  4. Mar 16, 2014 #3
    ok no problem
  5. Mar 16, 2014 #4

    Ray Vickson

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    I hope you did not mean what you wrote, which was
    [tex] \frac{dy}{dx} = 5(x^2 + 4x)^4 4x + 4, \text{ which } = 4 +5(x^2 + 4x)^4 4x [/tex]
    I hope you meant
    [tex]\frac{dy}{dx} = 5(x^2 + 4x)^4 (4x + 4) [/tex]
    If that is what you did mean, that is what you should write; note the parentheses.
  6. Mar 16, 2014 #5


    Staff: Mentor

    Use parentheses where they are needed.
    The right side should be 5u4 * (4x + 4)
    That last factor should be (4x + 4)
    No. I can't tell what you did here. How did you get 30 at the beginning of the right side?
  7. Mar 16, 2014 #6


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    In addition to what Ray noted, I can't figure out what you did to get the last two lines. You need to go back and review algebra.
  8. Mar 16, 2014 #7
    Ya sorry that is what I meant.

    can I simplify this further or can I leave it like that?
  9. Mar 16, 2014 #8


    Staff: Mentor

    You can factor 4 out of the 4x + 4 term, and put it with the 5 factor. Otherwise, that's about all you can do. For most purposes, leaving it in factored form is preferable to multiplying everything out.
  10. Mar 16, 2014 #9
    dy/dx=20(2x2+4x)*(4x) this what you mean
  11. Mar 16, 2014 #10


    Staff: Mentor

    No. Like vela said, you need to take some time to review algebra.

    Starting from here:
    dy/dx=5(2x2+4x)4 * (4x+4), factor 4 out of the last expression in parentheses, and combine that 4 with the leading 5. You did this, but the problem is that 4x + 4 ≠ 4*x. That seems to be what you're doing.
  12. Mar 17, 2014 #11



    Last edited: Mar 17, 2014
  13. Mar 17, 2014 #12


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    none of those. take your time, just using rules of arithmetic that you are certain about. for example, (4x+4) = 4*(x+1) right? So then what does the equation look like?
  14. Mar 17, 2014 #13
    I know this is probably very simple I just cant get it.


    Do I separate it out and treat 5(2x2+4x) from (4x+4)

  15. Mar 17, 2014 #14


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    yeah, almost. you forgot the first bit should be to the power of four. so it is 5(2x2+4x)4*4(x+1) And yes, it is fine to 'separate out'. In arithmetic, it is always OK to say a*(b*c)=a*b*c i.e. in this case 5(2x2+4x)4*(4x+4) = 5(2x2+4x)4*4(x+1)
  16. Mar 17, 2014 #15

    Thanks very much, frustrating that its so simple. thanks again
  17. Mar 17, 2014 #16


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    glad to have helped! yeah, I'm writing Makefiles at the moment, which should be a simple programming thing to do. But it's taking me ages! haha
  18. Mar 17, 2014 #17
    Best of luck.
  19. Mar 17, 2014 #18
    You are really not going to be able to do calculus unless you have good facility with algebra. You did not realize that you could combine the factors 5 and 4 to give 20. Your previous posts had several algebra errors in them. I implore you to please follow Vela's advice and review algebra.

  20. Mar 17, 2014 #19

    Thanks I have been looking over it today. It's just the work load of study, job and kids I can only do so much.

    I can multiply 5*4 to get 20. Can I do anything with what in the bracket?
  21. Mar 17, 2014 #20
    Yes, you can factor out a 2, and, when it comes out of the bracket, it becomes 24=16, which, when multiplied by the 20 becomes 320.

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