The Taylor Series of f(x) = exp(-x^2) at x = 0 is 1-x^2... Why is this? The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ... and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0??? Can someone please explain why it is 1-x^2???