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I know this is probably something I should be able to do by this stage but such is life.

- Thread starter jimmycricket
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- #1

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I know this is probably something I should be able to do by this stage but such is life.

- #2

pasmith

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\frac{\partial y}{\partial y''} = \frac{\partial y'}{\partial y''} = \frac{\partial x}{\partial y''}= 0 \\

\frac{\partial y''}{\partial y''} = 1.[/tex]

(b) To calculate the second derivative of that with respect to x, give y, y' and y'' their normal meanings: [tex]

\frac{dy''}{dx} = y''' \\ \frac{dy'}{dx} = y'' \\ \frac{dy}{dx} = y' \\ \frac{dx}{dx} = 1.[/tex]

- #3

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But I don't know how to extend this to find [tex]

\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/tex]

- #4

pasmith

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It's the same priniciple: to find [itex]\frac{d}{dx}(\frac{\partial F}{\partial y'})[/itex] you first calculate [itex]\frac{\partial F}{\partial y'}[/itex] by regarding y' as a variable independent of [itex]x[/itex] and [itex]y[/itex]. Then you differentiate the result with respect to [itex]x[/itex], regarding y' as the derivative of y with respect to x.

But I don't know how to extend this to find [tex]

\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/tex]

Example: let [itex]F(x,y,y') = x + yy' + y'^2[/itex]. Then [tex]

\frac{\partial F}{\partial y'} = y + 2y' \\

\frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = \frac{d}{dx}(y + 2y') = y' + 2y''.

[/tex]

It's the same principle with [itex]\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})[/itex]. Let [itex]F(x,y,y',y'') = x + y + y'y'' + e^{y''}[/itex]. Then [tex]

\frac{\partial F}{\partial y''} = y' + e^{y''} \\

\frac{d}{dx} \left(\frac{\partial F}{\partial y''} \right) = y'' + y'''e^{y''} \\

\frac{d^2}{dx^2} \left(\frac{\partial F}{\partial y''} \right) = y''' + y^{(4)}e^{y''} + (y''')^2 e^{y''}.[/tex]

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