Question

If f(x) = (lnx)^2, then f''(sqrt e)=
a. 0
b. 1
c. 2
d. e

Answer 1

The value of f''(sqrt e) is 1. (Option b)

Step-by-step explanation:

First derivative of f(x):

f'(x) = 2ln(x) * (1/x) = 2ln(x)/x

Second derivative of f(x):

f''(x) = (2/x^2) * (x - ln(x))

Evaluate f''(sqrt e):

f''(sqrt e) = (2/(sqrt e)^2) * (sqrt e - ln(sqrt e))

f''(sqrt e) = (2/e) * (sqrt e - 1/2)

f''(sqrt e) = (2sqrt e - 1) / e

f''(sqrt e) ≈ 1

Therefore, the value of f''(sqrt e) is approximately 1.