f(x) is a polynomial function, which is differentiable everywhere, including at x = 1.

The left-hand derivative is -1 and the right-hand derivative is 1. Since they are not equal, f(x) is not differentiable at x = 1.

The function |x - 1| is not differentiable at x = 1 due to a cusp.

The function f(x) = |x - 3| is not differentiable at x = 3 because it has a sharp corner.

f(x) is not differentiable at x = -2 and x = 2, but is differentiable everywhere else.

The function is a polynomial and is differentiable everywhere, hence nowhere.

The function is a polynomial and is differentiable everywhere. Thus, there are no points where it is not differentiable.

Setting f(1-) = f(1+) and f'(1-) = f'(1+) leads to a = 1 for differentiability.

Setting f(2-) = f(2+) and f'(2-) = f'(2+) leads to m = 1 for differentiability.

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

f'(x) = 2e^(2x). At x = 0, f'(0) = 2e^0 = 2.

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

Using the product rule, f'(x) = e^x * ln(x) + e^x/x. At x = 1, this simplifies to 0.

Using the product rule, f'(0) = e^0 * sin(0) + e^0 * cos(0) = 0 + 1 = 1.

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

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

f'(x) = sec^2(x). At x = π/4, f'(π/4) = sec^2(π/4) = 2.

Using the limit definition of the derivative, we find that f'(0) = 0, hence it is differentiable at x = 0.

f'(x) = 3x^2 - 6x. At x = 2, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

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

Set the left-hand limit equal to the right-hand limit and their derivatives at x = 2.

Setting the left-hand limit equal to the right-hand limit gives c = 1.

f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.

The function is a polynomial and is differentiable for all k, hence k can be any real number.

For f(x) to be differentiable at x = k, f'(k) must exist. Setting k = 1 makes f'(k) continuous.

f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

f'(x) = 2x + 2.

The function is a polynomial and is differentiable everywhere.

Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.