Posted by Ben on .
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,
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,
ahh thank you so much