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Derivative of constant

suvadip

Member
Feb 21, 2013
69
If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).
But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: derivative of constant

What does the definition of the derivative tell you?

edit: Sorry, I misread...have you tried integrating with respect to the independent variable?
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: derivative of constant

If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).
But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?
Another approach would be to use differentials:
$$\Delta y=f'(x)\, \Delta x = 0 \cdot \Delta x \, ...$$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: derivative of constant

If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).
But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?
The mean value theorem: if f is continuous and differentiable on [a, b] then there exist c in [a, b] such that [tex]f'(c)= \frac{f(b)- f(a)}{b- a}[/tex]. If f'(x)= 0 for all x, then for any a and b, [itex]\frac{f(b)- f(a)}{b- a}= 0[/tex] from which f(a)= f(b).