[tex] \frac{d}{dx} [e^{x}+e^{-\frac{x}{2}}]=\frac{d}{dx}e^{x}+\frac{d}{dx}(e^{-\frac{x}{2}})\\ \\ \frac{d}{dx}e^{x}=e^{x} \\ \\ \frac{d}{dx}(e^{-\frac{x}{2}})=e^{-\frac{x}{2}} \frac{d}{dx}[-\frac{x}{2}]=(-\frac{1}{2}\frac{d}{dx}x) e^{-\frac{x}{2}} \\ \\ \frac{d}{dx} [e^{x}+e^{-\frac{x}{2}}]=e^{x}-\frac{1}{2}e^{-\frac{x}{2}} [/tex]
si acum derivata a doua:
[tex] \frac{d}{dx}[e^{x}-\frac{1}{2}e^{-\frac{x}{2}}]=\frac{d}{dx}(e^{x})-\frac{1}{2}\frac{d}{dx}(e^{-\frac{x}{2}} ) [/tex]
[tex] =e^{x}-\frac{e^{-\frac{x}{2}}\frac{d}{dx}(-\frac{x}{2})}{2} [/tex]
[tex] =e^{x}-\frac{-\frac{1}{2}\frac{d}{dx}(x)e^{-\frac{x}{2}}}{2}= [/tex]
[tex] =e^{x}+\frac{e^-{\frac{x}{2}}}{4} =e^{x} +\frac{e^{-\frac{x}{2}}}{4}[/tex]
[tex] \it f'(x) = \left(e^x+e^{-\dfrac{x}{2}}\right)' = (e^x)' +\left(e^{-\dfrac{x}{2}}\right)' = e^x +\left(-\dfrac{x}{2}\right)' e^{-\dfrac{x}{2}} =
\\ \\ \\
= e^x -\dfrac{1}{2}e^{-\dfrac{x}{2}}
\\ \\ \\
f''(x) = \left(e^x -\dfrac{1}{2}e^{-\dfrac{x}{2}}\right)' = e^x-\dfrac{1}{2}\left(-\dfrac{x}{2}\right)'e^{-\dfrac{x}{2}}= e^x+\dfrac{1}{4} e^{-\dfrac{x}{2}} [/tex]