Chapter 9 Differentiation Miscellaneous Exercise 9 Solutions

Let y = x5 Differentiating w.r.t. x, we get

Let y = x-2 Differentiating w.r.t. x, we get

Let y = √x Differentiating w.r.t. x, we get

Let y = x√x ∴ y = Differentiating w.r.t. x, we get

Let y = ∴ y = Differentiating w.r.t. x, we get

Let y = 7x Differentiating w.r.t. x, we get

Demand, D = Rate of change of demand When price P = 2, Rate of change of demand, ∴ When the price is 2, the rate of change of demand is -3. ∴ Here, the rate of change of demand is negative demand would fall when the price becomes ₹ 2.

Given demand D = Now, marginal demand When P = 4 Marginal demand

Given, supply S = 2p3 + 5 Now, marginal supply ∴ When p = 5 Marginal supply = = 6(5)2 = 150 Here, the rate of change of supply with respect to the price is positive which indicates that the supply increases.

Total cost C = x2 + 4x + 4 Now. Average cost = = x + 4 + and Marginal cost = (x2 + 4x + 4) = (x2) + 4 (x) + (4) = 2x + 4(1) + 0 = 2x + 4 ∴ When x = 7, Marginal cost = = 2(7) + 4 = 14 + 4 = 18

Demand, D = Rate of change of demand = When price P = 3, Rate of change of demand, ∴ When price is 3, Rate of change of demand is -3.

Given, P = 20 + D – D2 Rate of change of price = = (20 + D – D2) = 0 + 1 – 2D = 1 – 2D Rate of change of price at D = 3 is = 1 – 2(3) = -5 ∴ Price is changing at a rate of -5, when demand is 3.

Total cost function C = 5x3 + 2x2 + 1 Average cost = = = 5x2 + 2x + When x = 4, Average cost = 5(4)2 + 2(4) + = 80 + 8 + = = Marginal cost = = (5x3 + 2x2 + 1) = 5 (x3) + 2 (x2) + (1) = 5(3x2) + 2(2x) + 0 = 15x2 + 4x When x = 4, marginal cost = = 15(4)2 + 4(4) = 240 + 16 = 256 ∴ The average cost and marginal cost at x = 4 are and 256 respectively.

Given, S = P2 + 9P – 2 ∴ The marginal supply is 23, at P = 7.

Given, cost C = x2 + 15x + 81 If marginal cost = average cost, then 2x + 15 = x + 15 + ∴ x = ∴ x2 = 81 ∴ x = 9 …..[∵ x > 0]