Wednesday, January 30, 2013
Attempt of an alternative proof of the Product Rule
| \[ \begin{aligned} \dfrac{d}{dx} \left(f(x) \cdot g(x) \right) &= \displaystyle\lim_{h \to 0} \frac{f(x+h) \cdot g(x+h)-f(x) \cdot g(x)}{h} \\ & = \displaystyle\lim_{h \to 0} \frac{\left(f(x) + h \cdot f'(x)\right)\left(g(x) + h \cdot g'(x)\right)-f(x) \cdot g(x)}{h} \\ &= \displaystyle\lim_{h \to 0} \frac{f(x) \cdot g(x) + h \cdot f'(x) \cdot g(x) + h \cdot g'(x) \cdot f(x) + h^2 \cdot f'(x) \cdot g'(x) - f(x) \cdot g(x)}{h} \\ &= \displaystyle\lim_{h \to 0} \left(f'(x) \cdot g(x) + g'(x) \cdot f(x) + h \cdot f'(x) \cdot g'(x) \right) \\ &= f'(x) \cdot g(x) + g'(x) \cdot f(x) \\ \end{aligned} \] |
There exists a subtle non-obvious step above. Can you see it ?
