# Question #64fc4

Dec 17, 2014

This will require the application of the chain rule.
First we see that $\frac{d}{\mathrm{dx}} {\left(b l o b\right)}^{2} = 2 \left(b l o b\right) \cdot \frac{\mathrm{db} l o b}{\mathrm{dx}}$ where blob is anything you can possibly imagine.
So we have $y ' = 2 \ln \left(1 + {e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(1 + {e}^{x}\right)\right)$

Next, $\frac{d}{\mathrm{dx}} \left(\ln \left(1 + {e}^{x}\right)\right) = \frac{1}{1 + {e}^{x}} \cdot {e}^{x}$
because the derivative of ${e}^{x} = {e}^{x}$

So putting everything together,

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

Finding y'' is more tricky because we will need to use quotient rule.
Remember, "Lowdeehi minus hideelow all over low squared."

$y ' ' = \frac{\left(1 + {e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{x}\right) - 2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{2 x}}{1 + {e}^{x}} ^ 2$
$\frac{d}{\mathrm{dx}} \left(2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{x}\right) = 2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{x} + \frac{2 {e}^{2 x}}{1 + {e}^{x}}$

$= \frac{\left(1 + {e}^{x}\right) \left[2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{x} + \frac{2 {e}^{2 x}}{1 + {e}^{x}}\right] - 2 \ln \left(1 + {e}^{x}\right) \cdot {e}^{2 x}}{1 + {e}^{x}} ^ 2$