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If

$f=f(g(x),h(x))$

Then I can easily demonstrate the chain rule:

$\frac{df}{dx}=\frac{\partial{f}}{\partial{g}}\cdot\frac{\partial{g}}{\partial{x}}+\frac{\partial{f}}{\partial{h}}\cdot\frac{\partial{h}}{\partial{x}}$

But what if

$f=f(x,g(x,t))$

Then it'll be wrong if I write the chain rule this way:

$\frac{\partial{f}}{\partial{x}}=\frac{\partial{f}}{\partial{x}}+\frac{\partial{f}}{\partial{g}}\cdot\frac{\partial{g}}{\partial{x}}$

because the $\frac{\partial{f}}{\partial{x}}$ in the left has a different meaning from the one in the right.

Then how do I express this equation?

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    Dear Voldemort: May I suggest an exercise? Given a function $f(x,y)$, do we have $\frac{\partial}{\partial x}\Big(f(y,x)\Big)=\frac{\partial f}{\partial x}(y,x)\quad?$ Hint: try $f(x,y)=x$.2011-08-20

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From Wolfram Alpha