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How would I calculate the total derivative using the chain rule with a multi-variable function? I need a total derivative with respect to $(t)$

$$R(t) = \max_x f(x^*(t, p), t)$$

Is this correct? $$ \frac{\partial R}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} \frac{\partial x}{\partial p} + \frac{\partial f}{\partial t} $$

Or do I need to extend it out?

$$ \frac{\partial R}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial x}\frac{\partial p}{\partial t} + \frac{\partial f}{\partial t} $$

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I'm not sure how you got to your first option, but your second option is almost right. The only thing missing is the partial of $x^*$ with respect to $p$ in the second term.

\begin{align} \frac{\mathrm d R}{\mathrm d t} &= \frac{\partial f}{\partial x^*} \frac{\mathrm d x^*}{\mathrm d t} + \frac{\partial f}{\partial t}\\ &= \frac{\partial f}{\partial x^*} \left( \frac{\partial x^*}{\partial t} + \frac{\partial x^*}{\partial p}\frac{\partial p}{\partial t} \right) + \frac{\partial f}{\partial t}\\ &= \frac{\partial f}{\partial x^*} \frac{\partial x^*}{\partial t} + \frac{\partial f}{\partial x^*}\frac{\partial x^*}{\partial p}\frac{\partial p}{\partial t} + \frac{\partial f}{\partial t} \end{align}

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  • $\begingroup$ Can you expand on the middle terms? $\endgroup$ – Amstell Sep 19 '17 at 5:42
  • $\begingroup$ I added some detail. $\endgroup$ – Epiousios Sep 19 '17 at 5:46
  • $\begingroup$ Ah, ok. So you need a partial effect of $p$ wrt $t$ taken from the derivative $dx/dt$. I was think I needed to expand out the middle term to account for that, but wasn't sure if the third term was necessary. This makes sense. Thanks! $\endgroup$ – Amstell Sep 19 '17 at 5:48

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