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Math Help - Applying chain rule

  1. #1
    Member VitaX's Avatar
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    Applying chain rule

    Find the derivative of the following:

    y = (2x - 5)(4 - x)^{-1}

    I got y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2} or y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}

    Correct?
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by VitaX View Post
    Find the derivative of the following:

    y = (2x - 5)(4 - x)^{-1}

    I got y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2} or y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}

    Correct?
    y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2} or y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2} looks wrong to me.

    You've got your product rule okay, but when you differentiated 2x-5 you got 2x rather than 2.

    A simple slip-up, it's so easy to do - other than that the technique is fine.
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by Matt Westwood View Post
    y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2} or y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2} looks wrong to me.

    You've got your product rule okay, but when you differentiated 2x-5 you got 2x rather than 2.

    A simple slip-up, it's so easy to do - other than that the technique is fine.
    ah so I did. Thanks for the help.
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  4. #4
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    Quote Originally Posted by VitaX View Post
    Find the derivative of the following:

    y = (2x - 5)(4 - x)^{-1}

    I got y' = 2x(4 - x)^{-1} + (2x - 5)(4 - x)^{-2} or y' = \frac{2x}{4 - x} + \frac{2x - 5}{(4 - x)^2}

    Correct?
    Personally, I'd use the quotient rule since y = \frac{2x - 5}{4 - x}.
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  5. #5
    Member VitaX's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Personally, I'd use the quotient rule since y = \frac{2x - 5}{4 - x}.
    Yah works a lot better and easier here. Final answer is y' = \frac{3}{(4 - x)^2}
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