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Math Help - Learning Derivatives

  1. #1
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    Learning Derivatives

    Beginner. Working on this problem: Use the definition of a derivative function to calculate f'(4) where f(x) = x + 3.
    Have made attempt using lim h->0 [f(x+h) - f(x)]/h but unsure if answer is correct. Need explanation. Help!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Yami Sorceress View Post
    Beginner. Working on this problem: Use the definition of a derivative function to calculate f'(4) where f(x) = x + 3.
    Have made attempt using lim h->0 [f(x+h) - f(x)]/h but unsure if answer is correct. Need explanation. Help!
     <br />
\lim_{h\to0}\frac{[(x+h)+3]-(x+3)}{h}

    Now simplify and take the limit? Got it?

    Hint: the limit is constant
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  3. #3
    Super Member Random Variable's Avatar
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    There is an equivalent definition that is sometimes useful:

    f'(c) = \lim_{x \to c} \frac {f(x) - f(c)}{x-c}

    so  f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x-4}  = \lim_{x \to 4} \frac {(x+3) -7}{x-4}  = \lim_{x \to 4}\frac {x-4}{x-4} = 1
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  4. #4
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    then ...

    OK, so for clarity I could write:

    lim h->0 (x+3)+h - (x+3) / h

    simplify:
    lim h->0 h/h = 1 ?

    Where does the f ' (4) fit in?
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Yami Sorceress View Post
    OK, so for clarity I could write:

    lim h->0 (x+3)+h - (x+3) / h

    simplify:
    lim h->0 h/h = 1 ?

    Where does the f ' (4) fit in?
     f'(4) = \lim_{h \to 0} \frac {f(4+h) - f(4)}{h}
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  6. #6
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    Thank you for the clear example.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Yami Sorceress View Post
    Where does the f ' (4) fit in?
    What you must understand is that f'(x)=1 for ALL x. So, f'(4)=1, f'(132,345,225)=1. All x. This makes sense right? because the slope of a line is constant right?
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