Answer to Question #280099 in Real Analysis for abc

Question

State suitable conditions and prove that (𝑓𝑔) ′ = 𝑓𝑔 ′ + 𝑓 ′𝑔.

Answer 1

Let us state the suitable conditions: if the functions $f$ and $g$ are differentiable in each point of their common domain, then the function $fg$ is differentiable in each point of this domain and $(𝑓𝑔)'= 𝑓𝑔' + 𝑓'𝑔.$

Let us prove that $(𝑓𝑔)'= 𝑓𝑔' + 𝑓'𝑔:$

$(f(x)g(x))' =\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)[g(x+\Delta x)-g(x)]+[f(x+\Delta x)-f(x)]g(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)[g(x+\Delta x)-g(x)]} {\Delta x}+ \lim\limits_{\Delta x\to 0}\frac{[f(x+\Delta x)-f(x)]g(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}f(x+\Delta x)\lim\limits_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)} {\Delta x}+ \lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}g(x) \\=f(x)g'(x)+f'(x)g(x).$