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Math Help - [SOLVED] Finding Derivatives

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] Finding Derivatives

    One Problem States:
    1. F(x) = 1/x , a=2
    By Using The definition of a derivative
    do i have to multiply the denominator and numerator by a fraction? Im getting a bit lost here and its not a good sign...

    there is another problem that says to use the alternate definition of a derivative where the numerator has f(x) - f(a), instead of (x+h). and to find the derivative at the point:

    2. (with the same problem above.)
    What Is The Difference and what accounts for that difference?
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  2. #2
    Moo
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    Hello
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    One Problem States:
    1. F(x) = 1/x , a=2
    By Using The definition of a derivative
    do i have to multiply the denominator and numerator by a fraction? Im getting a bit lost here and its not a good sign...
    \frac{f(a+h)-f(a)}{h}=\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac 12}{h} \quad (*)
    now, by noting that 2(h+2) is the common denominator, \frac{1}{2+h}-\frac 12=\frac{2}{2(2+h)}-\frac{2+h}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}

    (*)=\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2h(2+h)}=-\frac{1}{2(2+h)}

    now take the limit h -> 0 and you'll have the answer

    there is another problem that says to use the alternate definition of a derivative where the numerator has f(x) - f(a), instead of (x+h). and to find the derivative at the point:

    2. (with the same problem above.)
    What Is The Difference and what accounts for that difference?
    There is no difference


    f'(a)=\lim_{h \to 0} ~ \frac{f(a+h)-f(a)}{h}=\lim_{x \to a} ~ \frac{f(x)-f(a)}{x-a}

    from the first one to the second one, you just have to substitute x=a+h
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  3. #3
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    f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

    f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
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  4. #4
    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by Moo View Post
    Hello

    \frac{f(a+h)-f(a)}{h}=\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac 12}{h} \quad (*)
    now, by noting that 2(h+2) is the common denominator, \frac{1}{2+h}-\frac 12=\frac{2}{2(2+h)}-\frac{2+h}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}

    (*)=\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2h(2+h)}=-\frac{1}{2(2+h)}

    now take the limit h -> 0 and you'll have the answer


    There is no difference


    f'(a)=\lim_{h \to 0} ~ \frac{f(a+h)-f(a)}{h}=\lim_{x \to a} ~ \frac{f(x)-f(a)}{x-a}

    from the first one to the second one, you just have to substitute x=a+h
    Hola

    Ok, I Think i was getting confused with substituting a in finding the derivative (sounds weird, butt its how i word it)
    so i substitute what a is in f(a+h) and make it similar to what f(x), and f(a) is just substituting. ok i get it
    Gracias

    Sorry, Wrong Problem :[, but yes you are right
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  5. #5
    Moo
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    Hola

    Ok, I Think i was getting confused with substituting a in finding the derivative (sounds weird, butt its how i word it)
    so i substitute what a is in f(a+h) and make it similar to what f(x), and f(a) is just substituting. ok i get it
    Gracias

    but isn't there a difference? i tried #2 and i get 0.
    and the first one is 4.
    Hmmm the first one is -1/4 !

    and for #2, use the same method for \frac 1x-\frac 12
    you should get the same things.
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