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I want to calculate the derivative of a function with respect to, not a variable, but respect to another function. For example: $$g(x)=2f(x)+x+\log[f(x)]$$ I want to compute $$\frac{\mathrm dg(x)}{\mathrm df(x)}$$ Can I treat $f(x)$ as a variable and derive "blindly"? If so, I would get $$\frac{\mathrm dg(x)}{\mathrm df(x)}=2+\frac{1}{f(x)}$$ and treat the simple $x$ as a parameter which derivative is zero. Or I should consider other derivation rules?

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3 Answers

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$$\frac{dg(x)}{df(x)} = \frac{dg(x)}{dx} \cdot \frac{1}{f'(x)} = \frac{g'(x)}{f'(x)}$$

In your example,

$$g'(x) = 2f'(x) + 1 + \frac{f'(x)}{f(x)}$$

So:

$$\frac{dg(x)}{df(x)} = \frac{2f'(x) + 1 + \frac{f'(x)}{f(x)}}{f'(x)} = 2 + \frac{1}{f'(x)} + \frac{1}{f(x)}$$

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You can not. You have to derivate $f(x)$ as function.

$g'(x) = 2f'(x) + 1 + {f'(x) \over f(x)}$

EDIT: Sorry, That would make $dg(x) \over dx$, Deepak is right.

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You could if it were a function of $f(x)$ But it's not, due to the $x$ term.

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