Derivative of constant function - proof

[twoflower]

Hi all,

our professor wrote this proof (using the definition of derivative), that the derivative of constant function is 0:

<br /> f(x) \equiv a<br />

<br /> f^{&#039;}(b) = \lim_{h \rightarrow 0} \frac{f(b+h) - f(h)}{h} = \lim_{h \rightarrow 0} \frac{a - a}{h} = 0<br />

I'm not sure about the last step, because we have 0 denominator, don't we? Why isn't it here considered an indeterminate expression?

Thank you.

 

[Muzza]

Long-winded explanation: lim(h -> 0) 0/h = 0 is equivalent to

For all e > 0 there exists a d > 0 such that 0 < |h| < d => |0/h| < e.

Notice that since 0 < |h|, h is never equal to 0. Thus |0/h| is always 0 (for the values of h we are interested in!), so that 0 < e for all d. (So we can let d = 1, for example).

 

[twoflower]

Long-winded explanation: lim(h -> 0) 0/h = 0 is equivalent to

For all e > 0 there exists a d > 0 such that 0 < |h| < d => |0/h| < e.

Notice that since 0 < |h|, h is never equal to 0. Thus |0/h| is always 0 (for the values of h we are interested in!), so that 0 < e for all d. (So we can let d = 1, for example).

Thank you muzzo, I was confused with comparing it to the limits like

<br /> \lim_{x \rightarrow 0} \frac{\sin x}{x} = 0<br />

Where both nominator and denominator go to zero, but the difference is in that in this case we can't judge the result since both expressions are unlimitedly small, but not zero, unlike the proof, where nominator was legal zero. Am I right now?

 

[dextercioby]

Thank you muzzo, I was confused with comparing it to the limits like
<br /> \lim_{x \rightarrow 0} \frac{\sin x}{x} = 0<br />
Where both nominator and denominator go to zero, but the difference is in that in this case we can't judge the result since both expressions are unlimitedly small, but not zero, unlike the proof, where nominator was legal zero. Am I right now?

Yep,you're right.Differentiation should always be looked upon as process of evaluating a limit from a ratio.When the numerator is zero,as a difference between 2 identical constants,then the limit is identically zero,being a ratio bewteen zero and a positive/negative quantity,very,very,very small,but still nonnull.
The formula for the derivative of a constant is the most simple,i guess u know that.In all other cases,u really need to compute that limit involved by the definition.

Cheers and good luck!

Daniel.

 

[HallsofIvy]

A very important property of limits is this: if f(x)= g(x) for every x except c then they have the same limit at c (that's what you use over and over again for derivatives). Of course f(h)= 0/h and g(h)= 0 are exactly the same for h not equal to 0 so they have the same limit at 0: 0.

Another way of defining the derivative of a function (at 0₀) is "the slope of the tangent line to the graph y= f(x) at x= x₀". For any linear function, its graph is a straight line- it is its own "tangent line" and so its derivative is a constant: its slope. In particular, if f(x)= constant, the graph is a horizontal straight line and has slope 0.

 

[ankamiria]

Hi all,
if we want to prove the other way round, i.e. if the derivative is zero then f must be a constant function how could we do that? Any ideas?

 

[morphism]

Hi all,
if we want to prove the other way round, i.e. if the derivative is zero then f must be a constant function how could we do that? Any ideas?

That follows from the mean value theorem.

 

[chota]

Hi all,
if we want to prove the other way round, i.e. if the derivative is zero then f must be a constant function how could we do that? Any ideas?

INtegrals!

 

[HallsofIvy]

Just "integrals" isn't enough. If, for example, I know that df/dx= 2x, then I certainly know that f(x)= x² is one possible value for f. I presume that you know that f must be of the form x²+ C where "C" is a constant but HOW do you know that? How do you know that there isn't some very strange function, perhaps one that no one has ever written down, that has derivative 2x? You know it because of the mean value theorem as morphism said.

If F(x) and G(x) have the same derivative, F'(x)= G'(x) then (F- G)'= F'- G'= 0. Since the derivative of F- G is always 0, we can write [(F-G)(b)- (F-G)(a)]/(b-a)= (F-G)'(c)= 0 for any a and b. Then (F-G)(b)= (F-G)(a) for all a and b: F- G is a constant and F= G+ a constant.