# How do you find the derivative of

1. Jun 3, 2007

### AznBoi

How do you find the derivative of $$f(x)=5$$?? I got 0/0... but shouldn't there be no derivative then? I know that I can visualize the function f(x) because it is just a straight horizontal line and there is no slope at any point on the line... but how do you solve it algebraically so that it actually makes sense?

Last edited: Jun 3, 2007
2. Jun 3, 2007

### morphism

How did you get 0/0? The derivative of a constant function is 0. This is plausible because the slope isn't changing (and the derivative is the rate of change of slope). You can prove this pretty easily using the limit definition of a derivative.

3. Jun 3, 2007

### theperthvan

No, it's plausible because the slope is flat. What about a linear function, y=mx+c? The slope isn't changing, it's constant...but the deriv isn't 0.
So the bit in brackets should say "and the second derivative is the rate of change of slope"

4. Jun 4, 2007

### prasannapakkiam

D=capital delta
d=lowercase delta
y=5
therefore: y+Dy=5
therefore: y/Dx + Dy/Dx = 5/Dx
y=5, hence: y/Dx=5/Dx
therefore: Dy/Dx = 0
taking the limits: Dx-->0
therefore: dy/dx = 0 ==>

This is the algebraic proof you wanted...

5. Jun 4, 2007

### prasannapakkiam

I see that you got 0/0 probably using the l'hopital's rule. when you get (5-5)/h, it becomes 0/h which equals 0. You only take the limits once all simplifications are complete...

6. Jun 4, 2007

### VietDao29

I think you have messed up a little. A pure "0" differs from a fake "0" a lot. A pure "0" means that it's really 0, while a fake one means that it only tends to 0, it's not 0, it just tends towards 0.

You should notice that:
$$\lim_{x \rightarrow \infty} 0 x = 0$$, since 0 multiply by any number is 0, even if x is large.

Whereas:
$$\lim_{x \rightarrow \infty} \frac{1}{x} \times x = \lim_{x \rightarrow \infty} 1 = 1$$
Some may argue that when x tends to infinity, 1 / x tends to 0, so 0 * anything would be 0. But that's not true. 1 / x only tends to 0, it's not 0. So, you cannot claim that.

The same thing apply here:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{5 - 5}{h} = \lim_{h \rightarrow 0} \frac{0}{h}$$
Now the numerator is a pure 0, a real one. It's 0, so 0 divided by anything will give out 0, so f'(x) = 0.

Can you get it? :)

7. Jun 4, 2007

### cepheid

Staff Emeritus
I don't know if "real zero" and "fake zero" are very good terms to use. But what VietDao is describing is essentially the difference between "exactly zero" and "really small". Anyway, the good thing about his post is that he proves that the derivative of a constant function is zero using the *definition* of a derivative. It just occured to me to try using the power rule and see what happens:

$$f(x) = C$$

C is a constant.

$$\frac{d}{dx}C = \frac{d}{dx}Cx^0 = 0(Cx^{-1}) = 0$$

The result is zero for all x. Is this valid?

8. Jun 4, 2007

### C0nfused

I will just add that if you look at the definition of the limit (-> http://mathworld.wolfram.com/Limit.html ) it says "for any ε>0 there exists δ .... with 0<|x-x1|<δ" and not just |x-x1|<δ. So we don't care about the case x=x1. This actually shows the "fake" and "real" zero concept. If you use this definition and the definition of the derivative, then you can prove, as VietDao29 did, that the derivative of the constant funtion is 0

C=Cx^0 for x<>0. So, if you prove the rule for the derivative of x^n with n natural (zero in our case), you will still have to check the derivative at x=0 using the definition

Last edited: Jun 4, 2007