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.

  • $\begingroup$ product rule is still the product rule. $\endgroup$
    – Doug M
    Jun 7, 2017 at 2:00
  • $\begingroup$ @DougM I'm not sure how to apply product rule here because I'm still differentiatnig w.r.t. $t$, whilst one of the products in each sum have $\partial x$ or $\partial y$ in the "denominator" (so I'm not sure how to bypass this besides writing $\frac{\partial ^2 f}{\partial x \partial t}$ which I'm not sure if that makes any sense $\endgroup$ Jun 7, 2017 at 2:03

3 Answers 3


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: $$ \ddot{g} = f_x \ddot{x} + f_y \ddot{y} + \dot{x}^2 f_{xx} + \dot{x}\dot{y}(f_{xy}+f_{yx})+ \dot{y}^2 f_{yy}. $$

  • 3
    $\begingroup$ Thanks! I'm not sure (in your first equals line), how that bracket $\frac{d}{dt} \left( \frac{\partial f}{\partial x}\right)$ became $\left( \frac{dx}{dt}\frac{\partial}{\partial x}\frac{\partial f}{\partial x} + \frac{dy}{dt} \frac{\partial}{\partial y}\frac{\partial f}{\partial x}\right)$. How did you introduce the partial $x$'s (and partial $y$'s)? $\endgroup$ Jun 7, 2017 at 3:00
  • 5
    $\begingroup$ If I write $h(x)=\partial f/\partial x$, it's $$ \frac{dh}{dt} = \frac{dx}{dt} \frac{\partial h}{\partial x}+\frac{dy}{dt} \frac{\partial h}{\partial y}, $$ which we recognise as the chain rule. $\endgroup$
    – Chappers
    Jun 7, 2017 at 12:23
  • $\begingroup$ That made it so much more clear, thank you!!! $\endgroup$ Jun 7, 2017 at 12:38

$g'(t) = \frac{\partial g}{\partial x}\frac{d x}{dt} + \frac{\partial g}{\partial y}\frac{dy}{dt}$

$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}$


Another way to write it is as follows: let $\gamma (t) =(x(t),y(t))$. So, by chain rule, $$ \dot g(t)=(Df)_{\gamma (t)} \cdot \gamma(t) =\langle \vec \nabla f (\dot\gamma(t)), \dot \gamma (t)\rangle. $$ And applying it again, using the Leibniz rule, $$ \begin{align} \ddot{g} (t) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[ (Df)_{\gamma (t)} \cdot \dot\gamma(t) \right] \\ &= \frac{\mathrm{d}}{\mathrm{d}t}\left[ (Df)_{\gamma (t)} \right] \cdot \dot\gamma(t) + (Df)_{\gamma (t)}\cdot \ddot{\gamma} (t) \\ &= \left( (D^2 f)_{\gamma (t)} \cdot \dot \gamma (t) \right)\cdot \dot \gamma (t) + \vec \nabla f(\gamma(t)) \cdot \ddot{\gamma} (t)\\ &= \left( \left[ \begin{matrix} f_{xx} (\gamma (t)) & f_{xy}(\gamma (t)) \\ f_{yx} (\gamma (t)) & f_{yy}(\gamma (t)) \end{matrix}\right] \cdot \dot \gamma(t) \right) \dot \gamma(t) + f_x(\gamma(t))\ddot{x}(t)+f_y (\gamma(t))\ddot{y} (t) \\ &= f_{xx}(\gamma(t))\dot{x} ^2 +f_{yy} (\gamma(t))\dot{y}^2+2\dot{x}\dot{y} f_{xy} (\gamma(t)) + f_x(\gamma(t))\ddot{x}(t)+f_y (\gamma(t))\ddot{y} (t) . \end{align} $$ (Assuming second derivatives are equal, which is the case when $f$'s second-order mixed partial derivatives $f_{xy}(x,y)$ and $f_{yx}(x,y)$ exist and are continuous.) Re-writing the last equality while removing the "implicit" dependency on the varible $t$, we have $$ \ddot{g} = f_{xx}\dot{x} ^2 +f_{yy}\dot{y}^2+2\dot{x}\dot{y} f_{xy} + f_x\ddot{x}+f_y \ddot{y} . $$


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