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Need Help w/ Derivates

I should:

Poll closed Jun 6, 2006.
  1. Drop Calculus

    2 vote(s)
    16.7%
  2. Keep Trying

    10 vote(s)
    83.3%
  1. Jun 5, 2006 #1
    I'm having a really hard time grabbing a hold of this definition of a derivative concept.

    I know that the derivative of [tex]X-X^2 = 1-2X. [/tex], when solving w/ the power rule.
    But, I get really lost when I need to solve it using the definition of the derivitave. Can someone please explain to me how I get from the 1st step to the 2nd.

    [tex]F(X) = X - X^2 [/tex]

    1. [tex]F'(X) = \frac{F(X+H) - F(X)}{H}[/tex]

    2. [tex]= \frac{(X+H)^2 - X^2}{H}[/tex]


    The way I tried it, I just input [tex]X - X^2 [/tex] for X and I got [tex] (X - X^2 + H) - (X-X^2) [/tex]
     
    Last edited: Jun 5, 2006
  2. jcsd
  3. Jun 5, 2006 #2
    Nope... try to do that derivative again

    If f(x) is x-x^2, what is f(x+h) it's not x-x^2 + h, I will tell you that much, but it sounds like you have some issues with algebra as well. You applied the definition wrong, so give it another go.
     
  4. Jun 5, 2006 #3
    I fixed that first part sorry. lol
     
  5. Jun 5, 2006 #4
    Yeah, that's my problem, I don't know how to do that.
     
  6. Jun 5, 2006 #5
    Split up your function to make things much simpler. Let's say you have [tex]f(x)=x[/tex] and [tex]g(x)=-x^2[/tex]

    [tex]f'(x)=\lim_{h \to 0}\frac{(x+h)-(x)}{h}[/tex]

    [tex]g'(x)=\lim_{h \to 0}\frac{-(x+h)^2-(-x^2)}{h}[/tex]

    Work from there.
     
  7. Jun 5, 2006 #6
    Reading your first post, I'll try to clarify what I think is your problem. If you have [tex]f(x)=x^2[/tex], then something like f(2) is easy, right? You just plug in 2 for x and get 4. But what is f(a)? Do the same thing. f(a)=a^2. What is f(x+h) then? Plug in (x+h) for x. f(x+h)=(x+h)^2
     
  8. Jun 5, 2006 #7
    Why does the H get squared and negated?
     
  9. Jun 5, 2006 #8
    It's still not fixed on my screen. the derivative of x-x^2 should be 1-2x (I see it in the TeX code but for some reason it isn't showing up on your post).

    Let me elaborate a little. If f(x) = x-x^2, then f(x+h) = (x+h) - (x+h)^2, so expand this out. What is then f(x+h) - f(x)?
     
    Last edited by a moderator: Jun 5, 2006
  10. Jun 5, 2006 #9
    try refresh F5
     
  11. Jun 5, 2006 #10
    So i'm doing it backwards here?
     
  12. Jun 5, 2006 #11
    I don't understand what you mean by backwards. Look at my original post to see the correct method of using the limit definition of a derivative. I'm sorry I can't be more helpful right now.
     
  13. Jun 5, 2006 #12
    Ok, if i try it this way, I get: [tex] (x+h) + h - (x+h)^2 [/tex]
     
    Last edited: Jun 5, 2006
  14. Jun 5, 2006 #13
    You want to solve the limit of [tex] \frac { f(x+h) - f(x)} {h} [/tex] as h approaches 0, right? If f(x) = x - x^2, then f(x+h) - f(x) = (x+h) - (x+h)^2 - (x - x^2)). I'm a little concerned though as these problems you are having are algebraic and not calculus-based, as your thread suggests. Applying basic rules of algebra (up to factoring in the denominator to the expression) should fetch you the right answer.
     
    Last edited by a moderator: Jun 5, 2006
  15. Jun 5, 2006 #14
    You wouldn't happen to have a link for these algebra rules, would you?

    I understand how you plugged in [tex] x-x^2 [/tex] for f(x).
    but,
    I don' see how you got [tex](x+h) - (x+h)^2 [/tex] for [tex]f(x+h)[/tex] though.
     
  16. Jun 5, 2006 #15
    I'm going to repeat what was said here. But maybe seeing it worked out will help you a little bit more.

    You know the definition of the derivatie is:
    [tex] f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

    Now you seem to be caught up with the algebra of functions. Let me go through a few examples with you:

    Lets say you have a function:
    [tex] f(x)=x^2 [/tex]

    Now lets plug in a "cat" :)

    [tex] f(cat)=f(x=cat)=(cat)^2 [/tex]

    Now lets plug in [itex] a+b[/itex]

    [tex] f(a+b)=f(x=a+b)=(a+b)^2 [/tex]


    Now plugging in a cat is really not any harder then plugging what you need into the limit definition I gave above. But lets try an example:

    Lets create a function, say:
    [tex] f(x)=x^{1000} [/tex]

    You seem to be able to use the "non-limit rules" to find the derivative, so you know that:
    [tex] f'(x)=1000x^{999} [/tex]

    Right? That's pretty easy right.

    Now lets use the limit definition.

    [tex] f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \frac{(x+h)^{1000}-x^{1000}}{h} [/tex]

    Notice what we did.

    We just plugged in our values.

    Do you see that:
    [tex] f(x+h)=(x+h)^{1000} [/tex]

    You can think of it like this if you want.
    let [itex] x+h = cat [/tex]
    [tex] f(x)=x^{1000} [/tex]
    [tex] f(cat)=(cat)^{1000} [/tex]

    And we know that [itex] cat = x+h [/itex] so...
    [tex] f(cat=x+h)=(cat=x+h)^{1000} [/tex]
    or in a more popular way to write it:
    [tex] f(x+h)=(x+h)^{1000} [/tex]

    So really, you are just plugging in values.
     
  17. Jun 6, 2006 #16
    Thanks for your help.

    So I did: [tex] f(x) = x - x^2[/tex]

    [tex] f'(x) = \frac {x + h - (x + h)^2 - x - x^2}{h}[/tex]

    Now I tried to simplify but I get stuck or have done it wrong.

    [tex]= \frac {x + h - x^2 + 2xh + h^2 - x - x^2}{h}[/tex]

    [tex]= \frac {h - 2x^2 + 2xh + h^2}{h}[/tex] // I cancelled out the x and combined the x squared

    Is this right?
     
    Last edited: Jun 6, 2006
  18. Jun 6, 2006 #17

    TD

    User Avatar
    Homework Helper

    Watch out, at the end of the numerator you should have "-f(x)". With f(x) = x-x², this becomes -(x-x²) = -x+x².
     
  19. Jun 6, 2006 #18
    So, your saying I need to distribute the minus sign to f(x)?
     
  20. Jun 6, 2006 #19

    TD

    User Avatar
    Homework Helper

    Of course, when it says -f(x), f(x) is considered as a whole and should therefore be placed within parathesis. This gives -(x-x²) which simplifies to -x+x² and not -x-x².
     
  21. Jun 6, 2006 #20
    Ah ok, thanks for explaining that.
     
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