First, find f'(x) = 10x + 3. Then, f'(2) = 10(2) + 3 = 20 + 3 = 23.

f'(x) = 10x + 3; f'(1) = 10*1 + 3 = 13.

The first derivative f'(x) = 10x - 3, and the second derivative f''(x) = 10.

Using the chain rule, f'(x) = 2e^(2x).

f'(x) = e^x + 2x. Thus, f'(0) = e^0 + 2(0) = 1 + 0 = 1.

Find f'(x) = e^x - 2x. Setting f'(x) = 0 gives a local maximum at x = 1.

f'(x) = e^x; f'(0) = e^0 = 1.

f''(x) = e^x, thus f''(0) = e^0 = 1.

The second derivative f''(x) = d^2/dx^2(e^x) = e^x.

f'(x) = e^x and f''(x) = e^x. Therefore, f''(0) = e^0 = 1.

The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.

f'(x) = 1/x; f'(1) = 1, hence f is differentiable at x = 1.

f'(x) = 1/x. Therefore, f'(1) = 1/1 = 1.

f'(x) = 1/x. Therefore, f'(e) = 1/e.

f'(x) = 1/x.

f'(x) = (2x)/(x^2 + 1). At x = 1, f'(1) = (2*1)/(1^2 + 1) = 1.

Using the chain rule, f'(x) = (1/(x^2 + 1)) * (2x) = 2x/(x^2 + 1).

To find critical points, we set f'(x) = cos(x) - sin(x) = 0. This gives tan(x) = 1, leading to x = π/4 and x = 5π/4 in the interval [0, 2π].

Using the derivative rules, f'(x) = cos(x) - sin(x).

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

f(π/2) = sin(π/2) = 1.

Using the product rule, f'(x) = e^x(x^2 + 2x).

Using the product rule, f'(x) = e^x(x^2 + 2x).

Using the product rule, f'(x) = 2x * ln(x) + x.

Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.

f'(x) = 2x + 2, thus f'(1) = 2(1) + 2 = 4.

First derivative f'(x) = 2x + 2. Second derivative f''(x) = 2.

Calculating the derivative f'(x) = 2x + 2, we find f'(1) = 4.

f(-1) = (-1)^2 + 2*(-1) + 1 = 0. The function is a polynomial and is continuous everywhere, including at x = -1.

The vertex form is f(x) = (x + 1)^2, so the vertex is (-1, 0).