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

Aug 13, 2017

$f ' ' \left(x\right) = \frac{2 x {e}^{x} - 3 {e}^{x} - 2}{2 x + {e}^{x}} ^ 2$

#### Explanation:

If we rewrite using laws of logarithms, we get:

$f \left(x\right) = \ln {\left(x + {e}^{x}\right)}^{\frac{1}{2}} = \frac{1}{2} \ln \left(x + {e}^{x}\right)$

By the chain rule, we get:

$f ' \left(x\right) = \frac{1}{2 \left(x + {e}^{x}\right)} \cdot \left({e}^{x} + 1\right) = \frac{{e}^{x} + 1}{2 x + {e}^{x}}$

Now by the quotient rule, we get:

$f ' ' \left(x\right) = \frac{{e}^{x} \left(2 x + {e}^{x}\right) - \left(2 + {e}^{x}\right) \left({e}^{x} + 1\right)}{2 x + {e}^{x}} ^ 2$

$f ' ' \left(x\right) = \frac{2 x {e}^{x} + {e}^{2 x} - \left(2 {e}^{x} + {e}^{2 x} + {e}^{x} + 2\right)}{2 x + {e}^{x}} ^ 2$

$f ' ' \left(x\right) = \frac{2 x {e}^{x} + {e}^{2 x} - 2 {e}^{x} - {e}^{2 x} - {e}^{x} - 2}{2 x + {e}^{x}} ^ 2$

$f ' ' \left(x\right) = \frac{2 x {e}^{x} - 3 {e}^{x} - 2}{2 x + {e}^{x}} ^ 2$

Hopefully this helps!