/sci/ - Science & Math

>>11449978
I'll prove it for function in [math]\mathbb{R}[/math] instead of [math]\mathbb{R}^2[/math] for general convenience.
We have that [math] \frac{dfg}{dx} (0) = lim_{ \epsilon \rightarrow 0} \frac{fg( \epsilon) - fg(0)}{ \epsilon} = lim_{ \epsilon \rightarrow 0} \frac{fg( \epsilon)}{ \epsilon} [/math], where [math]g(0)=0[/math] by hypothesis.
We also have that [math]f(0) = \lim_{ \epsilon \rightarrow 0} f( \epsilon)[/math], by continuity, so [math] f(0) \frac{dg}{dx} (0) = \lim_{ \epsilon \rightarrow 0} f( \epsilon ) \frac{g( \epsilon) - g(0) }{ \epsilon } = \lim_{ \epsilon \rightarrow 0} \frac{ fg( \epsilon) }{ \epsilon} [/math].
>>11450386
Yes.