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Math Help - differential

  1. #1
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    differential

    i need general instructions for this one...


    thanks!
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  2. #2
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    Hello, Phoebe!

    Let: . y\:=\:\cosh(x^2 - 3x + 1) . where: . \cosh t \:= \:\frac{e^t - e^{-t}}{2}

    Find: . \frac{d^2y}{dx^2}

    If we are allowed to use these differentiation formula, it's easier.
    . . \frac{d}{dx}(\sinh x) \:=\:\cosh x\qquad\qquad\frac{d}{dx}(\cosh x) \:=\:\sinh x

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: . y \:=\:\cosh(x^2 - 3x + 1)

    Then: . \frac{dy}{dx}\:=\:\cosh(x^2-3x + 1)\cdot(2x - 3)

    And: . \frac{d^2y}{dx^2}\:=\:\sinh(x^2-3x+1)\!\cdot\!2 \:+ \:(2x-3)\!\cdot\!\cosh(x^2-3x+1)\!\cdot\!(2x - 3)

    . . . . \boxed{\frac{d^2y}{dx^2} \:= \:2\sinh(x^2-3x+1) + (2x-3)^2\cosh(x^2-3x+1)}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    They may expect us to use the Chain Rule on:

    . . y \:=\:\frac{1}{2}\left[e^{(x^2-3x+1)} + e^{-(x^2-3x+1)}\right] . . . ack!

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  3. #3
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    Thanks!

    now i understand that d^2y/dX^2 is just the second derivative...

    i think they expect us to use the exponent (e) with the chain rule... i can do it now.
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