Multivariate Chain Rule and second order partials
For the function
$g(t) = f(x(t),y(t))$, how would I find $g''(t)$ in terms of the first and second order partial derivatives of $x,y,f$? I'm stuck with the chain rule and the only part I can do is:
$$g'(t) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$
and one I differentiate again, I'm not sure how I can differentiate w.r.t $t$ with the partials involving $\frac{\partial f}{\partial x}$ etc.
2 Answers
$\begingroup$You can apply the chain rule again, as well as the product rule. Notice that $x,y$ are only functions of $t$, so the appropriate notation is $dx/dt$ and so on. Now, for example, $$ \frac{d}{dt} \left( \frac{\partial f}{\partial x} \frac{dx}{dt} \right) = \frac{\partial f}{\partial x} \frac{d}{dt} \left( \frac{dx}{dt} \right) + \frac{d}{dt} \left( \frac{\partial f}{\partial x} \right) \frac{dx}{dt} \\ = \frac{\partial f}{\partial x} \frac{d^2x}{dt^2} + \frac{dx}{dt} \left( \frac{dx}{dt} \frac{\partial}{\partial x} \frac{\partial f}{\partial x} + \frac{dy}{dt} \frac{\partial}{\partial y} \frac{\partial f}{\partial x} \right) \\ = \frac{\partial f}{\partial x} \frac{d^2x}{dt^2} + \left( \frac{dx}{dt} \right)^2 \frac{\partial^2 f}{\partial x^2} + \frac{dx}{dt} \frac{dy}{dt} \frac{\partial^2 f}{\partial y\partial x}, $$ as you successfully did for the first derivative. Equally, $$ \frac{d}{dt} \left( \frac{\partial f}{\partial y} \frac{dy}{dt} \right) = \frac{\partial f}{\partial y} \frac{d^2y}{dt^2} + \left( \frac{dy}{dt} \right)^2 \frac{\partial^2 f}{\partial y^2} + \frac{dy}{dt} \frac{dx}{dt} \frac{\partial^2 f}{\partial x\partial y} $$ so adding gives $$ g''(t) = \frac{\partial f}{\partial x} \frac{d^2x}{dt^2} + \frac{\partial f}{\partial y} \frac{d^2y}{dt^2} + \left( \frac{dx}{dt} \right)^2 \frac{\partial^2 f}{\partial x^2} + \frac{dx}{dt} \frac{dy}{dt} \left( \frac{\partial^2 f}{\partial y\partial x} + \frac{\partial^2 f}{\partial x\partial y} \right) + \left( \frac{dy}{dt} \right)^2 \frac{\partial^2 f}{\partial y^2} $$
The important thing to remember is that $\partial f/\partial x$ and friends are all still just functions, in the same way that $f$ itself is, albeit with rather more complicated symbols. Indeed, one can use the abbreviated notation $f_x$ (or sometimes $f_{,x}$) for $\partial f/\partial x$ and $\dot{x}=dx/dt$ (or sometimes $x'=dx/dt$), which makes the expression look a lot shorter, although perhaps not simpler: $$ \ddot{g} = f_x \ddot{x} + f_y \ddot{y} + \dot{x}^2 f_{xx} + \dot{x}\dot{y}(f_{xy}+f_{yx})+ \dot{y}^2 f_{yy}. $$
$\endgroup$ 3 $\begingroup$$g'(t) = \frac{\partial g}{\partial x}\frac{d x}{dt} + \frac{\partial g}{\partial y}\frac{dy}{dt}$
$g''(t) = $$(\frac{\partial}{\partial x})g'(t)\frac{d x}{dt} + (\frac{\partial}{\partial y})g'(t)\frac{d y}{dt}\\ \frac{\partial^2 g}{\partial x^2}(\frac{d x}{dt})^2 + 2\frac{\partial^2 g}{\partial x\partial y}(\frac{d x}{dt}\frac{d y}{dt}) +\frac{\partial^2 g}{\partial y^2}(\frac{d y}{dt})^2 + \frac{\partial g}{\partial x}\frac{d^2 x}{dt^2} + \frac{\partial g}{\partial y}\frac{d^2y}{dt^2}$
$\endgroup$
