Rates of change

dy/dx tells you how fast y changes when x changes. If y = x², then dy/dx = 2x: at x = 3, y is changing 6 times as fast as x.

Most rate questions are about time. Two quantities (say the radius r and the volume V of a balloon) both change with time t. You are told one rate and asked for the other. Link them with the chain rule:

dV/dt = (dV/dr) × (dr/dt)

Three steps every time:

• Formula: write the equation that connects the two quantities (V = 4/3 πr³, A = x², and so on).
• Differentiate both sides with respect to t.
• Substitute the given numbers, only at the end.

Example: a balloon's radius grows at 0.5 cm/s. How fast is its volume growing when r = 2 cm? V = 4/3 πr³, so dV/dt = 4πr² × dr/dt = 4π × 4 × 0.5 = 8π cm³/s.

A decreasing quantity has a negative rate. If a square's side x shrinks at 2 cm/s, dx/dt = –2. At x = 5: dA/dt = 2x × dx/dt = 2 × 5 × (–2) = –20, so the area falls at 20 cm²/s.

Do not substitute the numbers before differentiating: a value like r = 2 is true only at one moment, while the formula holds all the time.

Business words: if C(x) is the total cost of making x items, the marginal cost is dC/dx, roughly the cost of making one more item. Marginal revenue is dR/dx. Example: C(x) = 2x² + 5x + 100 gives marginal cost 4x + 5, which is ₹45 when x = 10.