Total derivative of a partial derivative

[jimmycricket]

Im doing a question on functionals and I have to use the Euler lagrange equation for a single function with a second derivative. My problem is I don't know how to evaluate $\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})$. Here y is a function of x, so $y'=\frac{dy}{dx}$.
I know this is probably something I should be able to do by this stage but such is life.

 

[pasmith]

(a) To calculate $\frac{\partial F}{\partial y''}$, treat y'' as a variable independent of x, y, and y':$$\frac{\partial y}{\partial y''} = \frac{\partial y'}{\partial y''} = \frac{\partial x}{\partial y''}= 0 \\<br /> \frac{\partial y''}{\partial y''} = 1.$$
(b) To calculate the second derivative of that with respect to x, give y, y' and y'' their normal meanings: $$\frac{dy''}{dx} = y''' \\ \frac{dy'}{dx} = y'' \\ \frac{dy}{dx} = y' \\ \frac{dx}{dx} = 1.$$

 

[jimmycricket]

Im a little confused by your answer. I know that $$\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial^2 F}{\partial y' \partial y}\frac{dy}{dx}+\frac{\partial^2F}{\partial y'^2}\frac{d^2y}{dx^2}+\frac{\partial^2 F}{\partial y'\partial x}$$
But I don't know how to extend this to find $$\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})$$

 

[pasmith]

Im a little confused by your answer. I know that $$\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial^2 F}{\partial y' \partial y}\frac{dy}{dx}+\frac{\partial^2F}{\partial y'^2}\frac{d^2y}{dx^2}+\frac{\partial^2 F}{\partial y'\partial x}$$
But I don't know how to extend this to find $$\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})$$

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

Example: let $F(x,y,y') = x + yy' + y'^2$. Then $$\frac{\partial F}{\partial y'} = y + 2y' \\<br /> \frac{d}{dx}\left( \frac{\partial F}{\partial y'} \right) = \frac{d}{dx}(y + 2y') = y' + 2y''.$$

It's the same principle with $\frac{d^2}{dx^2}(\frac{\partial F}{\partial y''})$. Let $F(x,y,y',y'') = x + y + y'y'' + e^{y''}$. Then $$\frac{\partial F}{\partial y''} = y' + e^{y''} \\<br /> \frac{d}{dx} \left(\frac{\partial F}{\partial y''} \right) = y'' + y'''e^{y''} \\<br /> \frac{d^2}{dx^2} \left(\frac{\partial F}{\partial y''} \right) = y''' + y^{(4)}e^{y''} + (y''')^2 e^{y''}.$$