1. ## (6)'=0

Stupid q, but...
What is the proof that (5)' =0?

f(x)=5
f'(x)=0

I see that the graph has NO SLOPE, but how to proove it arithmetically?

Many thanks!

2. ## Re: (6)'=0

5 is a constant. The derivative of any constant is 0.

If you want a more rigorous proof:

\displaystyle \displaystyle \begin{align*} f(x) &= 5 \\ &= 5x^0 \\ \\ f'(x) &= 0\cdot 5 x^{0 - 1} \\ &= 0x^{-1} \\ &= 0 \end{align*}

Or an even more rigorous proof:

\displaystyle \displaystyle \begin{align*} f(x) &= 5 \\ \\ f(x + h) &= 5 \\ \\ f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{5 - 5}{h} \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= \lim_{h \to 0} 0 \\ &= 0 \end{align*}

Ingenious!!!

4. ## Re: (6)'=0

Or, as you pointed out to begin with, this is a "linear function" and its derivative is just a constant- its slope. If f(x)= 5, then f(1)= 5 and f(2)= 5. The slope is $\displaystyle \frac{f(2)- f(1)}{2-1}= \frac{0}{1}= 0$.