1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of a Function

  1. Oct 15, 2007 #1
    If [tex]f(x) + x^{2}[f(x)]^{3} = 10[/tex] and f(1) = 2, find f '(1).

    I wish I could say I tried, but I don't know how to approach this problem..
    All I did was double check the formula worked by inputting 2 for f(x) and 1 for x

    Can someone tell me how to start this? And I'll go from there :)
  2. jcsd
  3. Oct 15, 2007 #2
    Take the derivative of both sides of the equation.
  4. Oct 15, 2007 #3
    And don't forget to use the chain rule on the second part. You'll end up with an equation that involves both f(x) and f'(x). At this point, sub in x=1 and f(1)=2 and solve for f'(x).
  5. Oct 15, 2007 #4
    Ok, so what I've done is:
    take derivative of both sides
    original = [tex]f(x) + x^{2}(f(x))^{3} = 10[/tex]
    so d/dx of f(x) is f'(x)
    now [tex]x^{2}(f(x))^{3}[/tex]'s derivative I'm a bit unsure of..

    What I did was use the product rule, so..
    [tex]x^{2}((f(x))^{3})' + (f(x))^{3}(x^{2})'[/tex]
    and [tex](f(x)^{3})' = 3(f(x))^{2}*f'(x)[/tex] right?
    it's [tex]x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x)[/tex]
    Then that means the whole equation becomes
    [tex]f'(x) + x^{2}(3(f(x))^{2}*f'(x)) + f(x)^{3}(2x) = 0[/tex]
    In which I plugged in the numbers
    [tex]f'(1) + (1)^{2}(3(2)^{2}*f'(1)) + (2)^{3}(2(1)) = 0[/tex]
    [tex]f'(1) + 12*f'(1)) + 16 = 0[/tex]
    which means
    [tex](13)f'(1)) = -16[/tex]
    [tex]f'(1) = -16/13[/tex]???
    Does that make sense? Is it wrong? i have a feeling it is :/
  6. Oct 15, 2007 #5
    looks right.

    if you wanted to not deal with f(x) you could change f(x)=y and do implicit differentiation like you've been doing before and solve for y' and then change y to f(x).
  7. Oct 15, 2007 #6
    f(x)+x^(2)*f(x)^(3)//use product rule for the second term along with the chain rule
    f'(x)[1+3x^(2)]=-2xf(x)^(3)//moved 2xf(x)^(3) to other side and factored a f'(x)
    Last edited: Oct 15, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Derivative of a Function
  1. Deriving function (Replies: 13)

  2. Functional Derivatives (Replies: 27)