f ' (x) = 8 x⁷ - 2 (using shortcut for polynomial functions) Thus:

lim_h->0 (f ' (x₀ + h) - f ' (x₀)) / h

= lim_h->0 ((8(2 + h)⁷ - 2 - (8(2)⁷ - 2)) / h

= lim_h->0 ((8(h⁷ + 14h⁶ + 84h⁵ + 280h⁴ + 540h³ + 672h² + 448h = 128) - 2 - (8(128) - 2)) / h

= lim_h->0 ((8h⁷ + 112h⁶ + 672h⁵ + 2240h⁴ + 4480h³ + 5376h² + 3584h + 1024 - 2) - (1024 - 2)) / h

= lim_h->0 (8h⁷ + 112h⁶ + 672h⁵ + 2240h⁴ + 4480h³ + 5376h² + 3584h + 1024 - 2 - 1024 + 2)) / h

= lim_h->0 (8h⁷ + 112h⁶ + 672h⁵ + 2240h⁴ + 4480h³ + 5376h² + 3584h) / h

= lim_h->0 (h(8h⁶ + 112h⁵ + 672h⁴ + 2240h³ + 4480h² + 5376h + 3584)) / h

= 8(0)⁶ + 112(0)⁵ + 672(0)⁴ + 2240(0)³ + 4480(0)² + 5376(0) + 3584

= 3584 = d²/dx² [x⁸ -2x + 3] |_x=2

= 200 (900 - 60t - t²)

= 180,000 - 12,000t + 200t²

= 200t² - 12,000t + 180,000

Thus, Q ' (t) = 400t - 12,000

When t = 10, Q ' (t) = Q' (10) = 400(10) - 12,000

= 4,000 - 12,000

= -8,000

Therefore, at t = 10, ten minutes after the tank begins to drain, the tank is emptying at a rate of 8,000 gallons/minute.

(b) At t = 0, Q (t) = 200(30-0)² = 200 * 900 = 180,000 At t = 10, Q (t) = 200(30-10)² = 200 * 400 = 80,000

Average rate of drain = {delta}Q(t)/{delta}t = (Q(t)_f - Q(t)₀) / (t_f - t₀)

= (180,000 - 80,000) / (10 - 0) = 100,000 / 10

= 10,000 gallons/minute