I have a function $f(x)$ defined as follows:
$$ f(x)=\frac{1-g(x)}{1-xg(x)}. $$
Because $f$ contains the function $g(x)$, I guess you could say $f$ is a function of $g(x)$ and $x$.
Given this, a mathematician says
$$ \frac{\partial}{\partial x}f=\frac{\partial f}{\partial g}\frac{\partial g}{\partial x} + \frac{\partial f}{\partial x}. $$
I'm not following his reasoning. For one, I would have thought that $\frac{\partial}{\partial x}f=\frac{\partial f}{\partial x}$. So I'm not sure how that other term is fitting in here.
Can anyone explain this?
Assume that your $f$ depends as $$f=f(x,g)=\dfrac{1-g}{1-xg}.$$
Now, by use of chain's rule of vector analysis, we compute : $$\dfrac{df}{dt} = \dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial g}\dfrac{dg}{dt},$$ and this is for any parameter $t$. But if we take $t=x$ we get $$\dfrac{df}{dx} = \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial g}\dfrac{dg}{dx},\qquad (*)$$
Now, the partial derivatives of $f$ are $$\dfrac{\partial f}{\partial x}=\dfrac{-(1-g)(-g)}{(1-xg)^2}=\dfrac{(1-g)g}{(1-xg)^2}$$ and $$\dfrac{\partial f}{\partial g} =\dfrac{-(1-xg)-(1-g)(-x)}{(1-xg)^2} =\dfrac{x-1}{(1-xg)^2},$$ then, by subbing them into $(*)$, we receive: $$\dfrac{df}{dx} = \dfrac{(1-g)g}{(1-xg)^2} + \dfrac{x-1}{(1-xg)^2} \dfrac{dg}{dx}.\qquad (1) $$
Also there is an alternative anyone can do.
If $f(x)=\dfrac{1-g(x)}{1-xg(x)}$ then $$f'(x)=\dfrac{d}{dx}\left(\dfrac{1-g(x)}{1-xg(x)}\right),$$ which can be developed employing elementary derivative's properties.
The result would be $$\dfrac{df}{dx}=\dfrac{(1-g)g+(x-1)g'}{(1-xg)^2}.\qquad (2)$$
Both $(1)$ and $(2)$ coincide.