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
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