Finding the derivative of a constant function at a given point.

[Deepak singh]

I am very new to such online forums so please forgive me for any mistake i might have made.


f(x)=5

This is a real valued function and defined on all points.

what is the Derivative of f(x) at x=0?

so the way i solved is below...

f'(x)= f (x+h)-f(x)/h when lim h tends to 0

because x=0

It gives us f'(x)= 5-5/0 = 0/0 which is undefined

i don't know where did i go wrong. as

I myself intuitively feel that it should be 0 as this function's slope is 0 everywhere.

Can someone please help how come it has 2 answers?

Thank you for you help in advance.

 

[HallsofIvy]

Where you went wrong is in not taking the limit. Just setting h= 0 does NOT give the limit because the "difference quotient" is not continuous in h. In fact, the difference quotient, \(\displaystyle \frac{f(x+ h)- f(x)}{h}\) always has "0" in the denominator which is why we need the "limit". For f(x)= 5, a constant function, f(x+h)- f(x)= 5- 5= 0 for all h. Therefore \(\displaystyle \frac{f(x+h)- f(x)}{h}= \frac{0}{h}= 0\) for all h. The limit of "0", as h goes to 0, is 0.

 

[Deepak singh]

Sorry i didn't get the point very well.

i understood that

f'(x)= 0/h where h tends to 0 but how f'(x)=0/h=0 i didn't quite get it.

Do you mean that we can't have the h=0 as h cant be exactly equal to zero as this is denominator if that happens the value of quotient wont be defined.
so as this is not exactly 0 may some number very close zero..

let say

h=.00000000000000000000001

it but not zero so 0/h = 0/.00000000000000000000001= 0

and f'(x)=0 for all points as the denominator becomes equal to 0 always.

please tell me if this is what you mean
Thank you in advance!

 

[HallsofIvy]

No, that is not what I mean. It is true that you cannot set h equal to 0 because it is in the denominator but it is not true that you set h equal to some small number like =.00000000000000000000001.

You have to take the limit as h goes to 0. You should have learned what "limits" are and how to find "limits" before you learned the derivative.