Question

Use differential to approximate the following : 

(i) \(\sqrt{0.0037}\)

(ii) log_e (10.02), when log_e10 = 2.3026

Answer 1

(i) \(\sqrt{0.0037}\)

Let y = √x, x = 0.0036, y = 0.06

∆x = 0.0037 – 0.0036 = -0.001

∵ y = √x​​​​

Diff. w.r.t. x

Hence, approximate value of  \(\sqrt {0.0037}\) is 0.0608.

(ii) log_e (10.02), when log_e10 = 2.3026

Let y = log_e x

Where x = 10, ∆x = 0.02

and x + ∆x = 10.02

∵ y = log_e x

Diff. w.r.t. x,

From equation (i),

y + ∆y = log_e (x + ∆x)

log_e x + ∆y = log_e (x + ∆x)

⇒ log_e 10 + 0.002 = log_e (10.02)

⇒ log_e (10.02) = 2.3026 + 0.002

⇒ log_e (10.02)= 2.3046

Hence, approximate value of loge (10.02) is 2.3046.