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

f'(x) = cos(x); f'(π/2) = cos(π/2) = 0.

f'(x) = sec^2(x); f'(0) = sec^2(0) = 1.

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

The derivative f'(x) = d/dx(tan(x)) = sec^2(x).

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

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.

Using the power rule, f'(x) = 3x^2 - 6x + 4.

Using the power rule, f'(x) = 3x^2 - 8x + 6.

The determinant of an upper triangular matrix is the product of its diagonal elements, which is 1.

The determinant is calculated as \( 3*4 - 2*1 = 12 - 2 = 10 \).

The determinant of the identity matrix is always 1.

The determinant is calculated as \( 2*4 - 1*3 = 8 - 3 = 5 \).

Using the determinant formula, we find it equals 10.

The determinant evaluates to 0.

The determinant is \( 2*4 - 3*1 = 8 - 3 = 5 \).

The determinant is given by the formula \( ad - bc \).

This is the identity matrix, and its determinant is 1.

The determinant evaluates to 0 as the third row can be expressed as a linear combination of the first two.

det = (1*5) - (2*3) = 5 - 6 = -1.

For the parabola y^2 = 4px, here 4p = -8, so p = -2. The directrix is given by x = -p, which is x = 2.

Distance = √[(7-3)² + (1-4)²] = √[4 + 9] = √13 ≈ 3.6, closest option is 4.

Distance = √[(5-2)² + (7-3)²] = √[3² + 4²] = √[9 + 16] = √25 = 5.

Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2) = |3(1) + 4(2) - 12| / sqrt(3^2 + 4^2) = |3 + 8 - 12| / 5 = 1.

Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2) = |2*3 + 3*4 - 6| / sqrt(2^2 + 3^2) = |6 + 12 - 6| / sqrt(13) = 12 / sqrt(13).

The characteristic polynomial is det(A - λI) = (2-λ)(2-λ) - 1 = λ^2 - 4λ + 3 = 0, giving eigenvalues 1 and 3.

Equation of circle: (x-h)² + (y-k)² = r² => (x-2)² + (y+3)² = 5².

The equation y = mx + c represents a family of straight lines where m is the slope and c is the y-intercept.

The slope m = (4-2)/(3-1) = 1. Using point-slope form: y - 2 = 1(x - 1) gives y = x + 1.