# Math Help - How to evaluate the second total derivative of a 2D function

1. ## How to evaluate the second total derivative of a 2D function

suppose i have F(x,y), and i want to evaluate d2F. is the following correct?

$dF = \frac{{\partial F}}{{\partial x}}dx + \frac{{\partial F}}{{\partial y}}dy$

$d^2F = \frac{{\partial}}{{\partial x}} dF + \frac{{\partial}}{{\partial y}}dF$

$\frac{{\partial}}{{\partial x}} dF = \frac{{\partial^2 F}}{{\partial x^2}}dx^2 + \frac{{\partial^2F}}{{\partial y \partial x}}dydx$

$\frac{{\partial}}{{\partial y}} dF = \frac{{\partial^2 F}}{{\partial y^2}}dy^2 + \frac{{\partial^2F}}{{ \partial x \partial y}}dxdy$

$\frac{{\partial^2F}}{{\partial x \partial y}} dxdy = \frac{{\partial^2F}}{{\partial y \partial x}} dydx \Rightarrow$

$d^2F = \frac{{\partial^2 F}}{{\partial x^2}} dx^2 + \frac{{\partial^2F}}{{\partial y^2}} dy^2 + 2 \frac{{\partial^2 F}}{{\partial y \partial x}}dydx$

i don't believe i ever studied higher-order total derivatives in college, and now that i have a need to make use of it, i am unsure if $2 \frac{{\partial^2 F}}{{\partial y \partial x}}dydx$ should be $\frac{{\partial^2 F}}{{\partial y \partial x}}dydx$

edit:
thanks for the clue-in on LaTeX plato.

2. Originally Posted by method
if someone could tell me how to do the sub/superscript formatting without the use of html sub / sup tags, please let me know!
You can learn to use LaTeX.
Typing $$\frac{{\partial ^2 f}}{{\partial x^2 }}$$ gives $\frac{{\partial ^2 f}}{{\partial x^2 }}$ .

3. aah. excellent. i'll spruce the OP up for legibility.

do you happen to know the answer to the original Q ?

4. let $f(t)=f\big(x(t), y(t)\big)$

then $\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$

$\frac{d^{2}f}{dt^{2}} = \frac{d}{dt}\Big(\frac{\partial f}{\partial x} \frac{dx}{dt}\Big)$ $+ \frac{d}{dt} \Big(\frac{\partial f}{\partial y} \frac{dy}{dt} \Big)$ (sum rule)

$= \frac{dx}{dt}\frac{d}{dt} \Big(\frac{\partial f}{\partial x}\Big) + \frac{\partial f}{\partial x} \frac{d^{2}x}{dt^{2}} + \frac{dy}{dt}\frac{d}{dt} \Big(\frac{\partial f}{\partial y}\Big) + \frac{\partial f}{\partial y} \frac{d^{2}y}{dt^{2}}$ (product rule)

$= \frac{dx}{dt}\Bigg(\frac{\partial}{\partial x} \Big(\frac{\partial f}{\partial x}\Big)\frac{dx}{dt} + \frac{\partial}{\partial y} \Big(\frac{\partial f}{\partial x}\Big) \frac{dy}{dt} \Bigg)$ $+ \frac{\partial f}{\partial x} \frac{d^{2}x}{dt^{2}} + \frac{dy}{dt}\Bigg(\frac{\partial}{\partial x} \Big(\frac{\partial f}{\partial y}\Big)\frac{dx}{dt} + \frac{\partial}{\partial y} \Big(\frac{\partial f}{\partial y}\Big) \frac{dy}{dt} \Bigg)+ \frac{\partial f}{\partial y} \frac{d^{2}y}{dt^{2}}$ (first total derivative)

$= \frac{dx}{dt}\Big(\frac{\partial^{2} f}{\partial x^{2}}\frac{dx}{dt} + \frac{\partial^{2} f}{\partial y \partial x} \frac{dy}{dt} \Big)$ $+ \frac{\partial f}{\partial x}\frac{d^{2}x}{dt^{2}} + \frac{dy}{dt}\Big(\frac{\partial^{2} f}{\partial x \partial y}\frac{dx}{dt} + \frac{\partial^{2}f}{\partial y^{2}} \frac{dy}{dt} \Big)+ \frac{\partial f}{\partial y} \frac{d^{2}y}{dt^{2}}$

$= \frac{\partial^{2} f}{\partial x^{2}}\Big(\frac{dx}{dt}\Big)^{2} + \frac{\partial^{2} f}{\partial y \partial x} \frac{dx}{dt} \frac{dy}{dt} + \frac{\partial f}{\partial x}\frac{d^{2}x}{dt^{2}} + \frac{\partial^{2} f}{\partial x \partial y} \frac{dy}{dt} \frac{dy}{dt} + \frac{\partial^{2} f}{\partial y^{2}}\Big(\frac{dy}{dt}\Big)^{2} + \frac{\partial f}{\partial y}\frac{d^{2}y}{dt^{2}}$

and assuming the mixed derivatives are equal

$= \frac{\partial^{2} f}{\partial x^{2}}\Big(\frac{dx}{dt}\Big)^{2} + 2 \ \frac{\partial^{2} f}{\partial x \partial y} \frac{dx}{dt} \frac{dy}{dt} + \frac{\partial f}{\partial x}\frac{d^{2}x}{dt^{2}} + \frac{\partial^{2} f}{\partial y^{2}}\Big(\frac{dy}{dt}\Big)^{2} + \frac{\partial f}{\partial y}\frac{d^{2}y}{dt^{2}}$

5. that's a lotta LaTeX!

thanks! that does answer the question.