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Simple example:

$f(x) = x \cdot g(x) \cdot h(x)$

$g(x) = 5x^2$

$h(x) = 2x^3$

First, derive $f$ by $x$, then substitute $g$ and $h$:

$\frac{\partial f}{\partial x} = g(x) \cdot h(x) = 10x^5$

And now the other way round, substitute and then derive:

$\frac{\partial f}{\partial x} = \frac{\partial (10 x^6)}{\partial x} = 60 x^5$

What's wrong here? Why does it matter in which order I subsitute known functions in a partial derivative? What is the meaning of this?

2 Answers 2

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The problem is that you treated $g(x)$ and $h(x)$ as constants. They're not, they depend on $x$. If you take that into account and write $g'(x)$ and $h'(x)$ for their derivatives with respect to $x$, respectively, you'll get the same result regardless of when you substitute for them.

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    Thanks all of you. I chose this answer as the correct one in combination with Willies comment which solved my problem fully ("independent" was the keyword, I was not aware of this).2012-11-02
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It doesn't appear that you're applying the product rule correctly in the first case.

If you apply the product rule on $f(x)=xg(x)h(x)$ you should get

$\frac{df}{dx} = xg(x)h'(x)+xg'(x)h(x)+g(x)h(x).$

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    Another way of looking at it is $\frac{df}{dx}$ is the change in $f$ as a result of a small change in $x$. If $x$ changes, then as you have written it, so too does $g(x)$ and $h(x)$. Therefore, we must account for the effects of changes of $g(x)$ and $h(x)$ as a consequence of small changes of $x$.2012-11-02