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Math Help - Derivative of a Constant help

  1. #1
    Senior Member topher0805's Avatar
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    Derivative of a Constant help

    Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

    Find h'(3) for:

    (a) h(x) = 2f(x) - 5g(x)

    (b) h(x) = f(x)g(x)

    (c) h(x) = f(x)/g(x)

    (d) h(x) = g(x)/(2 + f(x))

    As far as I know, the derivative of a constant is always 0. So when I plug in the values of f(x) and g(x) to find h(3) I always get a constant, and so the derivative of that must be zero, right?
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  2. #2
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    Hello, topher0805!

    You're reading the problem incorrectly.

    . . h'(3) means: .find h'(x), then substitute  x = 3


    Given that: . f(3) = \text{-}5,\;\;f'(3) = \text{-}2,\;\;g(3) = \text{-}2,\;\;g'(3) = 3

    Find h'(3) for:

    (a)\;\;h(x) \:= \:2\!\cdot\!f(x) - 5\!\cdot\!g(x)

    We have: . h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)

    \text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}
    - . - . . h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19

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  3. #3
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    Quote Originally Posted by topher0805 View Post
    Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

    Find h'(3) for:

    [snip]

    (d) h(x) = g(x)/(2 + f(x))
    Using the quotient rule:

    h'(x) = \frac{g'(x) (2 + f(x)) - f'(x) g(x)}{(2 + f(x))^2}


    \Rightarrow h'(3) = \frac{g'(3) (2 + f(3)) - f'(3) g(3)}{(2 + f(3))^2}


    \Rightarrow h'(3) = \frac{(3) (2 - 5) - (-2) (-2)}{(2 - 5)^2} = .....
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by topher0805 View Post
    Given that f(3) = -5, f '(3) = -2, g(3) = -2, and g'(3) = 3.

    Find h'(3) for:

    (a) h(x) = 2f(x) - 5g(x)
    f(x) and g(x) are not constants, they are functions. So:

    Initial equation of problem a
    h(x)=2f(x)-5g(x)

    then
    \frac d{dx}h(x)=\frac d{dx}[2f(x)-5g(x)]

    simplify
    \frac d{dx}h(x)=\frac d{dx}[2f(x)]-\frac d{dx}[5g(x)]

    Use Newtonian Notation
    h\prime (x)=2f\prime (x)-5g\prime (x)

    then
    h\prime (3)=2f\prime (3)-5g\prime (3)

    and substitute values
    h\prime (3)=2(-2)-5(3)

    simplify
    h\prime (3)=-4-15 = -19


    Can you do the rest?



    edit: dang, I got doubly owned >.<
    Last edited by angel.white; February 22nd 2008 at 09:27 PM.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, topher0805!

    You're reading the problem incorrectly.

    . . h'(3) means: .find h'(x), then substitute  x = 3



    We have: . h'(x) \;=\;2\!\cdot\!f'(x) - 5\!\cdot\!g(x)

    \text{Then: }\;h'(3) \;=\;2\!\cdot\!\underbrace{f'(3)}_{\downarrow} - 5\!\cdot\!\underbrace{g'(3)}_{\downarrow}
    - . - . . h'(3) \;=\;\;\;2(\text{-}2) \;- \;5(3) \;\;=\;\;-4-15 \;\;=\;\;-19

    How did you line that up so well without arrays?
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  6. #6
    Math Engineering Student
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    You can also use the aligned environment:

    \begin{aligned}<br />
  h'(3) &= 2 \cdot \underbrace {f'(3)}_ \downarrow  - \,5 \cdot \underbrace {g'(3)}_ \downarrow\\<br />
   &= ~~2( - 2) ~~- ~\,5(3)\\<br />
   &=  - 4 - 15\\<br />
   &=  - 19.<br />
\end{aligned}

    -----

    Soroban uses the following: color beige, it's the perfect stealth to do what Soroban did.
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  7. #7
    Super Member angel.white's Avatar
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    Makes sense, of course, if MHF ever gets more skins, it won't be nearly so effective.
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