# Derivative of constant

##### Member
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

Staff member
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?

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#### Ackbach

##### Indicium Physicus
Staff member
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
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 $$f'(c)= \frac{f(b)- f(a)}{b- a}$$. 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).