f(x)=(1+x) what is f'(x) when x=0
2 answers
the derivative of f(x)=x+1 is f'(x)=1 so your answer would just be 1
so 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 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