DE PRODUCT REGEL

te bewijzen:

$\displaystyle p(x)=f(x)\cdot g(x)\rightarrow p'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x)$

bewijs:

Stel $ p(x)=f(x)\cdot g(x)$

$ \frac{\bigtriangleup p}{\bigtriangleup x}=\frac{p(x+h)-p(x)}{h}=\frac{f(x+h)\c...
...\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, }$ =

$ \frac{f(x+h)\cdot g(x+h)-f(x)\cdot g(x+h)+f(x)\cdot g(x+h)-f(x)\cdot g(x)}{h}=\frac{g(x+h)\cdot \left(f(x+h)-f(x)\right)+f(x)\cdot \left(g(x+h)-g(x)\right)}{h}$=

$ \frac{g(x+h)\cdot \left(f(x+h)-f(x)\right)}{h}+\frac{f(x)\cdot \left(g(x+h)-g(x)\right)}{h}=g(x+h)\cdot \frac{f(x+h)-f(x)}{h}+f(x)\cdot \frac{g(x+h)-g(x)}{h}$

$ \underset{h\rightarrow 0}{lim}\left(g(x+h)\cdot \frac{f(x+h)-f(x)}{h}+f(x)\cdot \frac{g(x+h)-g(x)}{h}\right)=g(x)\cdot f'(x)+f(x)\cdot g'(x)=p'(x)$