Please!

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!

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- Jul 25th 2013, 12:45 AM #1

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- Jul 25th 2013, 12:58 AM #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:

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

- Jul 25th 2013, 03:19 AM #3

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- Jul 25th 2013, 05:41 AM #4

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