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!

Calculate derivatives

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose f and g are functions. Assume that f is differentiable and f(1) = 2, f'(1)=4 and that g(x) = x^4 -x+1. Use this information to calculate:

    (fg)'(1)
    (f o g)'(1)


    2. Relevant equations



    3. The attempt at a solution

    I'm unsure of what this wants me to do Does it want me to multiply the derivatives of f and g together and calculate it at 1 or does it want me to multiply f and g then find the derivative and plug in 1.

    For the 2nd one I found the derivative of g(x) to be 4x^3 -1 plugged a 1 into that and got 3 now i have f'(3) = ? how can i know that when they only gave me f(1) and f'(1) ?
     
    Last edited: Oct 7, 2007
  2. jcsd
  3. Oct 7, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    I would guess that (fg)(x) means f(x) times g(x), but you should check your book or course notes or whatever is relevant for the definition you are supposed to be using.

    For the second part, (f o g)(x) usually means f(g(x)), which can be differentiated by the chain rule. You should find that you do not need the value of f'(3).
     
  4. Oct 7, 2007 #3
    Does "product rule" and "chain rule" ring a bell?
    Product rule: [tex](fg)'=f'g+fg'[/tex]
    and
    Chain rule: [tex] (f \circ g)'(x) = f'(g(x)) g'(x)[/tex]

    Just a clarification of notation: [tex](f \circ g)(x)[/tex] is also written as [tex]f(g(x))[/tex] and is read as "[tex]f[/tex] of [tex]g[/tex] of [tex]x[/tex]". It simply means you first apply the [tex]g[/tex] function to [tex]x[/tex] and then the [tex]f[/tex] function to the resultant. The [tex]'[/tex] indicates the first derivitive of the function.
     
  5. Oct 7, 2007 #4
    ah ok its not (fg)x tho its (fg)'(x)

    and I don't know the equation for f(x) it doesn't give it to you so how can i find the derivative of

    f(x^4-x+1)
    would it be
    (3x^3 - 1)*(f(x^4 - x+1))
     
  6. Oct 7, 2007 #5
    You don't need the equation for [tex]f(x)[/tex].

    Ok, lets do the first one first. Start off with the product rule:
    [tex](fg)'=f'g+f'[/tex] which can be written more formally as:
    [tex](f(x)g(x))'=f'(x)g(x) + f(x)g'(x)[/tex]

    The values of [tex]f'(1)[/tex] and [tex]f(1)[/tex] are given, and the values for [tex]g'(1)[/tex] and [tex]g(1)[/tex] can be calculated by plugging [tex]1[/tex] into the [tex]g(x)[/tex] function. And then differentiate [tex]g(x)[/tex] to find [tex]g'(1)[/tex]. Plug in the resulting values into the above equation and you will obtain your solution for the first part.

    Finding the derivitive of [tex]g(x)[/tex] should be quite simple.

    To do the second part, similarly just plug in your knowns.
     
    Last edited: Oct 7, 2007
  7. Oct 7, 2007 #6
    gotcha thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculate derivatives
Loading...