Introduction
d/dx
is another notation for the derivative.
\frac{d}{dx}f(x)=f'(x)
We will prove the following:
\frac{d}{dx} \left[f(x)g(x) \right] =f(x)g'(x)+f'(x)g(x)
Proof
\begin{align*}
\frac{d}{dx}f(x)g(x)&=\lim_{h \to 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h} \\\\
&=\lim_{h \to 0}\frac{f(x+h)g(x+h) \c1{+ f(x+h)g(x) - f(x+h)g(x)} - f(x)g(x)}{h} \\\\
&=\lim_{h \to 0}\frac{\c1{f(x+h)[g(x+h) - g(x)]} + \c1{g(x)[f(x+h) - f(x)]}}{h} \\\\
&=\c1{\lim_{h \to 0}}\frac{f(x+h)[g(x+h) - g(x)]}{h} + \c1{\lim_{h \to 0}}\frac{g(x)[f(x+h) - f(x)]}{h} \\\\
&=\lim_{h \to 0}\c1{f(x+h)}\frac{g(x+h) - g(x)}{h} + \lim_{h \to 0}\c1{g(x)}\frac{f(x+h) - f(x)}{h} \\\\
&=\c1{\lim_{h \to 0}}f(x+h) \cdot \c1{\lim_{h \to 0}}\frac{g(x+h) - g(x)}{h} + \c1{\lim_{h \to 0}}g(x) \cdot \c1{\lim_{h \to 0}}\frac{f(x+h) - f(x)}{h} \\\\
&=f(x) g'(x) + g(x) f'(x)
\end{align*}