f(x)=(1+x) what is f'(x) when x=0
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
To find the derivative of a function, we can use the power rule. The power rule states that if we have a function of the form f(x) = x^n, where n is a constant, then the derivative with respect to x is given by f'(x) = n*x^(n-1).
In this case, our function is f(x) = (1+x), which can be rewritten as f(x) = 1 + x^1. Applying the power rule, we can find the derivative of f(x) as follows:
f'(x) = d/dx(1 + x) = d/dx(1) + d/dx(x)
= 0 + 1
= 1
Therefore, the derivative of f(x) = (1+x) is f'(x) = 1.
Now, we want to find f'(x) when x=0. To do this, we substitute x=0 into our derivative equation:
f'(0) = 1
So, f'(x) is equal to 1 when x=0.