Posted by **Ben** on Tuesday, November 8, 2011 at 7:34pm.

So from a previous post I asked,

"f(x)=(1+x) what is f'(x) when x=0?"

and someone replied,

"the derivative of f(x)=x+1 is f'(x)=1 so your answer would just be 1"

Now my question is why!? it doesn't matter what x is? f'x when x=238423904 will also be 1? For some reason I thought that you would have to plug in 0 for x.. like you've got f(x) = (1+x) and when x = 0, f(0) = (1+0) = 1. So f'(0) would = 0. Can anybody help me rationalize why this is wrong? thanks in advance

- calculus -
**Steve**, Wednesday, November 9, 2011 at 4:40am
The derivative measures how much f(x) changes when x changes. That is called the slope. A (suitably smooth) curved graph has a tangent line at every point, whose slope is the derivative of the function at that value of x.

Now, the function f(x) = 1+x has a graph which is a straight line. The slope never changes. So, the graph of the derivative is just the line y=1.

So, yes, when x=238423904, f'(x) is still just 1. You can't plug in different values for x, because f'(x) does not depend on x in any way.

- calculus -
**Ben**, Wednesday, November 9, 2011 at 2:21pm
ahh thank you so much

## Answer This Question

## Related Questions

- Calculus - This is a problem we did previously on a test; my answer was marked ...
- Social studies? - Please ignore my previous post. I prematurely hit the enter ...
- CALC 1 - Suppose that h is a function with h'(x) = xe^x. Find all intervals ...
- calculus - what is the derivative? - Let f(x) = 2x^2 + 1 a. Find the derivative ...
- General - Please understand that I am not upset with anyone or whatever I just ...
- Chemistry - I just wanted to clarify that although Cody did respond to my ...
- Calculus - Find the derivative: h(x)=7x/sqrt(5-2x) Using the chain rule...I ...
- Calculus - derivatives - Okay, I want to find the derivative of (x^x)^(x^x)... ...
- Chemistry. - hi again, thanks to the previous help i got the first part of the ...
- try - I guess my question that I posted does not come up I would like to answer ...

More Related Questions