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Math Help - Could use some Derivation help, please

  1. #1
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    Could use some Derivation help, please

    Okay so I just started with derivation.. it's pretty damn fun but the Swedish standard math book sucks completly because they can't give a decent example even if their life depended on it.
    Anywho here is problem.

    Calculate f`(4) by using numeric derivation, if;

    a) f(x) = 3x^4 -350x + 129

    b) f(x) = x / (x^2 + 1)

    c) f(x) = 6.5 * 10^(0,0086x)


    On a) I tried to use the basic lim h > 0, it did work but there was so many damn numbers It's like made for you to screw up.
    f(4+h) - f(4)
    3(4+h)^4 - 350(4+h) + 129

    You see it right? (4+h)^4 is pure pain to keep track of.. is there an easier way for me to simplify this so the calculation won't get so long?

    On b) I got stuck completly.
    f(4+h) -f(4) = ((4+h) / ((4+h)^2 +1))) - 4 / (4^2 + 1)

    Could someone help me expand this and give me some derivation tips?

    P.S Thx to your help I got a B+ on my last math thest which I've never had before! Thanks ALOT!
    D.S
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  2. #2
    Moo
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    Hello,

    D'you know the general formulas for derivations ?

    The derivative of a sum is the sum of the derivatives etc...



    Or you can do it simply :

    f(4+h)-f(4)=3(4+h)^4-350(4+h)+129-3(4)^4+350(4)-129 =3\left((4+h)^4-4^4 \right)-350(4+h-4)=3\left((4+h)^4-4^4 \right)-350h

    Let's look what is \left((4+h)^4-4^4 \right)

    (4+h)^4=((4+h)^2)^2
    4^4=(4^2)^2=16^2

    So \left((4+h)^4-4^4 \right)=a^2-b^2 with a=(4+h)^2=16+8h+h^2 and b=16

    \left((4+h)^4-4^4 \right) =a^2-b^2=(a-b)(a+b) = (16+8h+h^2-16)(16+8h+h^2+16)=(8h+h^2)(32+8h+h^2)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by Hanga View Post
    Okay so I just started with derivation.. it's pretty damn fun but the Swedish standard math book sucks completly because they can't give a decent example even if their life depended on it.
    Anywho here is problem.

    Calculate f`(4) by using numeric derivation, if;

    a) f(x) = 3x^4 -350x + 129

    b) f(x) = x / (x^2 + 1)

    c) f(x) = 6.5 * 10^(0,0086x)


    On a) I tried to use the basic lim h > 0, it did work but there was so many damn numbers It's like made for you to screw up.
    f(4+h) - f(4)
    3(4+h)^4 - 350(4+h) + 129

    You see it right? (4+h)^4 is pure pain to keep track of.. is there an easier way for me to simplify this so the calculation won't get so long?

    On b) I got stuck completly.
    f(4+h) -f(4) = ((4+h) / ((4+h)^2 +1))) - 4 / (4^2 + 1)

    Could someone help me expand this and give me some derivation tips?

    P.S Thx to your help I got a B+ on my last math thest which I've never had before! Thanks ALOT!
    D.S
    Here I will go thorough the definition of the derivative for the first one and see if it helps...ok we have f(x)=3x^4-350x+129....and we have f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}...so we have f'(x)=\lim_{h \to 0}\frac{3(x+h)^4-350(x+h)+129)-(3x^4-350x+129)}{h}...expanding all but the first term we get f'(x)=\lim_{h \to 0}\frac{3(x+h^4)-350x-350h+129-3x^4+350x-129}{h}...simplifying we get f'(x)=\lim_{h \to 0}\frac{3(x+h)^4-350h+3x^4}{h}./...now expanding we get f'(x)=\lim_{h \to 0}\frac{3x^4+12x^3h+18x^2h^2+12xh^3+3h^4-350h-3x^4}{h}..combining like terms we get f'(x)=\lim_{h \to 0}\frac{12x^3h+18x^2h^2+12xh^3+3h^4-350h}{h}....now by splitting the top apart we get f'(x)=\lim_{h \to 0}\bigg[12x^3+18x^2h+12xh+3h^3-350\bigg]...now since all the terms with h go to zero just the terms with just x are left...therfore f'(x)=12x^2-350...and then f'(4)=12\cdot{4^3}-350=418
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  4. #4
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    Yeah I can see that there is no shortcut in this :P
    Could you guys help me with b) please? I'd really like to see how you solve that one.
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  5. #5
    Moo
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    Ok :P

    f(x) = \frac{x}{x^2 + 1}

    \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h} ?

    First of all, do you know the other formula :

    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} ?
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  6. #6
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    Quote Originally Posted by Moo View Post
    Ok :P

    f(x) = \frac{x}{x^2 + 1}

    \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h} ?

    First of all, do you know the other formula :

    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} ?
    Yeah I know the
    \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}

    I do not know this one
    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} ?[/quote]
    Simply because there is no example on how to execute this in my lousy **** book...

    enlighten me S.V.P

    I tired using this;

    (f(4+h) - f(4) / h)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    I'll let moo take the work

    Quote Originally Posted by Hanga View Post
    Yeah I know the
    \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}

    I do not know this one
    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} ?
    Simply because there is no example on how to execute this in my lousy **** book...

    enlighten me S.V.P

    I tired using this;

    (f(4+h) - f(4) / h)[/quote]
    but the second one is the alternate form of the deriavative that says f'(x) evaluated at c is f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    Simply because there is no example on how to execute this in my lousy **** book...

    enlighten me S.V.P

    I tired using this;

    (f(4+h) - f(4) / h)
    but the second one is the alternate form of the deriavative that says f'(x) evaluated at c is f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}[/quote]

    Here is what I don't undestand. How do I use the f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}<br />
Can you please show me an example?

    For me this looks like it's always going to be 1....
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  9. #9
    Moo
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    Quote Originally Posted by Hanga View Post
    Yeah I know the
    \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}

    I do not know this one
    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} ?
    Simply because there is no example on how to execute this in my lousy **** book...

    enlighten me S.V.P

    I tired using this;

    (f(4+h) - f(4) / h)
    What a pity >_<

    Then you should know the one i gave you, it'll be easier to calculate i think.
    You can get this formula by changing variables :
    h -> x-a

    That is to say h+a -> x



    Forget it , i'm not sure it's more simple
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  10. #10
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    Quote Originally Posted by Moo View Post
    What a pity >_<

    Then you should know the one i gave you, it'll be easier to calculate i think.
    You can get this formula by changing variables :
    h -> x-a

    That is to say h+a -> x



    Forget it , i'm not sure it's more simple
    Please give me an example using
     f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}


    what is the difference between f(a) and a? What is the diffirence between f(x) and x?

    If I take f(x) - f(a) / x - a

    wont the sum alwas become 1?

    f(x) - f(2) / x - 2 = 1!?

    please just give me an example so I can see how you use this formula
    f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}

    PLEASE!
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  11. #11
    Moo
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    Ok, i went through it.

    I'm sorry, but it's far less complicated to write it than type it

    If you have questions about it, just tell me which line !
    Attached Thumbnails Attached Thumbnails Could use some Derivation help, please-ex.jpg  
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  12. #12
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    thanks mate, thats awsome!

    I understand how to do something like that now

    Viva le france
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  13. #13
    Moo
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    You're welcome

    And f(x) is completely different from x !
    f(x) is the image of x by the function f. It will not necessarily return the same value )
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