Question

What is the second derivative of `f(x)= ln sqrt(x+e^x)`?

Answer 1

`f''(x) = (2xe^x - 3e^x - 2)/(2x + e^x)^2`

Explanation:

If we rewrite using laws of logarithms, we get:

`f(x) = ln(x + e^x)^(1/2)= 1/2ln(x + e^x)`

By the chain rule, we get:

`f'(x) = 1/(2(x + e^x)) * (e^x + 1) = (e^x + 1)/(2x + e^x)`

Now by the quotient rule, we get:

`f''(x) = (e^x(2x + e^x) - (2 + e^x)(e^x + 1))/(2x + e^x)^2`

`f''(x) = (2xe^x + e^(2x) - (2e^x + e^(2x) + e^x + 2))/(2x + e^x)^2`

`f''(x) = (2xe^x + e^(2x) - 2e^x - e^(2x) - e^x - 2)/(2x + e^x)^2`

`f''(x) = (2xe^x - 3e^x - 2)/(2x + e^x)^2`

Hopefully this helps!