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Derivative question

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    If d/dx (f(2x^4)) = 8x^5

    Calculate f'(x)


    2. Relevant equations



    3. The attempt at a solution

    Well , we have been only doing derivatives for 2-3 class and its new for me.

    From what I understand in this question is.

    You have a function f(x) = something
    The derivative of this function when you plug in 2x^4 gives you 8x^5
    With that i suppose I should be able to find the equation of f(x) and then give f'(x) but I really don't know how to actually do this.
     
  2. jcsd
  3. Oct 28, 2007 #2
    well are you finding the derivative of 2x^4 because if you are then it's not 8x^5. It's 8x^3. You use the power rule, which is x^n = nx^n-1

    edit: If what your saying is correct then just work backwards from the power rule...
     
    Last edited: Oct 28, 2007
  4. Oct 28, 2007 #3
    Thats something diffrent the question is already saying

    d/dx (f(2x^4) = 8x^5
     
  5. Oct 28, 2007 #4
    I understand but if the derivative of a function is 8x^5 can you find the function?
     
  6. Oct 28, 2007 #5
    I guess am suppose to find the function but I don't know how. Am I suppose to take the power rule and use it in reverse or something of the sort
     
  7. Oct 28, 2007 #6
    well i told you the power rule and you know what the derivative is going to be can't you work backwards to get to your function?
     
  8. Oct 28, 2007 #7
    ok so assuming we have to workd backwards the function of 8x^5 would be 1.6x^6
     
  9. Oct 28, 2007 #8
    I don't think so b/c 6 times 1.6 will not give you 8...
     
  10. Oct 28, 2007 #9
    oh opps took the 5 instead yeah.

    so 1.33 x ^ 6
     
  11. Oct 28, 2007 #10
    okay so you have the function f(2x^4)= 4/3 x^6. Now you need to find f(x)
     
  12. Oct 28, 2007 #11
    there could be numerous possibilities... if you are working backwards....because how will you find constants that equal to zero? they could be anything.
    check the back of the book for the answer and see what it really wants.

    To me I think it wants.
    f(2x^4) = 8x^5 //compute this
    Then take the derivative of 8(2x^4)^5.
     
  13. Oct 28, 2007 #12
    well its a work we do on the internet and the answers won't be submitted before some date so I guess I'll ask the teacher or other students next class but I still want to keep on trying for now
     
  14. Oct 28, 2007 #13
    I agree with pooface b/c the way your really going about the problem just seems complicated for a class that has just started derivatives. Maybe the d/dx was a typo or you misread it. Then you would just do what pooface said by finding that derivative
     
  15. Oct 28, 2007 #14
    well in the whole 19 questions only 1-3 are like this. the others a basic , just apply the rules
     
  16. Oct 28, 2007 #15
    Which is why i doubt they want you to do this complicated thing they probably just wanted you to sub the x with 2x^4 into 8x^5 and find the derivative
     
  17. Oct 28, 2007 #16
    well if I correctly did the derivative of 8(2x^4)^5 it gives me 45(2x^4)*8x^3

    but the web thing says the answer is incorrect . wether I simpligy or not
     
  18. Oct 28, 2007 #17
    My teacher does like putting in nasty problems sometimes tho. Maybe thats it ? sometimes he writes soemthing partlially and says you will know how next week.
     
  19. Oct 28, 2007 #18
    wouldn't your derivative be 5120x^19 not whatever you got
     
  20. Oct 28, 2007 #19
    no that is not correct. (to evil bunny)

    8*5 is 40 not 45.

    Where is the exponent 5-1?

    you can multiply the exponents out first if you like.
     
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