# What is the second derivative of f(x)=e^x + e^(-x / 2 ?

Mar 12, 2018

${d}^{2} / {\mathrm{dx}}^{2} f \left(x\right) = f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(f ' \left(x\right)\right)$

the first derivative by using the chain rule is
$\frac{d}{\mathrm{dx}} f \left(x\right) = {e}^{x} - {e}^{- \frac{x}{2}} / 2$

then take the derivative of that function

therefore,
again using the chain rule on the second element,

$\frac{d}{\mathrm{dx}} f ' \left(x\right) = {e}^{x} + {e}^{- \frac{x}{2}} / 4$

$f ' \left(x\right) =$
${e}^{x} + {e}^{- \frac{x}{2}} / 4$
=

Mar 12, 2018

$\textcolor{b l u e}{{e}^{x} + \frac{1}{4} {e}^{- \frac{1}{2} x}}$

#### Explanation:

Since the second derivative is the derivative of the first derivative, we start by finding:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x} + {e}^{- \frac{x}{2}}\right)$

Re-writing:

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

We know $\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} = {e}^{x}$

We need to use the Chain Rule for ${\left({e}^{x}\right)}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x} + {\left({e}^{x}\right)}^{- \frac{1}{2}}\right) = {e}^{x} - \frac{1}{2} {\left({e}^{x}\right)}^{- \frac{3}{2}} \cdot {e}^{x}$

$= {e}^{x} - \frac{1}{2} {e}^{- \frac{3}{2} x + x} = {e}^{x} - \frac{1}{2} {e}^{- \frac{1}{2} x}$

We now differentiate this again:

Re-writing as before:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x} - \frac{1}{2} {\left({e}^{x}\right)}^{- \frac{1}{2}}\right) = {e}^{x} + \frac{1}{4} {\left({e}^{x}\right)}^{- \frac{3}{2}} \cdot {e}^{x}$

$= {e}^{x} + \frac{1}{4} {e}^{- \frac{3}{2} x + x} = {e}^{x} + \frac{1}{4} {e}^{- \frac{1}{2} x}$

$\therefore$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left({e}^{x} + {e}^{- \frac{x}{2}}\right) = \textcolor{b l u e}{{e}^{x} + \frac{1}{4} {e}^{- \frac{1}{2} x}}$