Question

Use differential to approximate the following :  

(i) (3.968)^3/2

(ii) (32.15)^1/5

Answer 1

(i) (3.968)^3/2

Let y = x^3/2, x = 4

y = (4)^3/2 = (22)^3/2 = 23 = 8

∆x = 3.968 – 4 = -0.032

∵ y = x^3/2

Diff. w.r.t. x

Hence, approximate value of (3.968)^3/2 is 7.904.

(ii) (32.15)^1/5

Let y = x^1/5, x = 32, y = (32)^1/5 = 2,

∆x = 32.15 – 32 = 0.15

∵ y = x^1/5

Diff. w.r.t. x

Hence, approximate value of (32.15)^1/5 is 2.001875.