Commission Brokerage and Discount — Practice Quiz

(2x + 3)/(x² + 3x)

Using chain rule for logarithm: d/dx[ln(u)] = (1/u) × du/dx. Step 1: Let u = x² + 3x, then du/dx = 2x + 3. Step 2: d/dx[ln(x² + 3x)] = (1/(x² + 3x)) × (2x + 3) = (2x + 3)/(x² + 3x). The key is to multiply by the derivative of the inside function.

-2/(x - 1)²

Quotient rule: d/dx[u/v] = (u'v - uv')/v². Step 1: u = x + 1, v = x - 1. Step 2: u' = 1, v' = 1. Step 3: f'(x) = [(1)(x-1) - (x+1)(1)]/(x-1)² = [x-1-x-1]/(x-1)² = -2/(x-1)². Remember to be careful with the signs in quotient rule.

2/(1 + 4x²)

For inverse tangent: d/dx[tan⁻¹(u)] = (1/(1 + u²)) × du/dx. Step 1: Let u = 2x, then du/dx = 2. Step 2: d/dx[tan⁻¹(2x)] = (1/(1 + (2x)²)) × 2 = (1/(1 + 4x²)) × 2 = 2/(1 + 4x²). Don't forget to square the entire inside function.

d/dx[x⁵] = 5x⁴, d/dx[cos(x)] = -sin(x), d/dx[ln(x)] = 1/x

Step 1: d/dx[x⁵] = 5x⁴ (power rule) ✓. Step 2: d/dx[cos(x)] = -sin(x) ✓. Step 3: d/dx[ln(x)] = 1/x ✓. Step 4: d/dx[eˣ] = eˣ, not xeˣ⁻¹ ✗. The exponential function eˣ is special because its derivative is itself.