Derivatives of Exponential Function y = 5/2(e^x/5 + e^-x/5)

[kimmy_koo51]

If y = 5/2(e^x/5 + e^-x/5), prove that y" = y/25.

 

[stapel]

If y = 5/2(e^x/5 + e^-x/5)...

Does the above mean any of the following?

. . . . .y = (5/2) (e^x/5 + e^-x/5)

. . . . .y = 5 / [2 (e^x/5 + e^-x/5)]

. . . . .y = (5/2) [(e^x) / 5 + (e^-x) / 5]

. . . . .y = 5 / {2 [(e^x) / 5 + (e^-x) / 5] }

Or did you mean something else?

When you reply, please include all the steps you have tried so far, starting with your first and second derivatives, showing what you got for y / 25, and what your thoughts were when you compared your second derivative to y / 25.

Thank you.

Eliz.

Edit: Ne'mind; the full solution is provided below.

 

[soroban]

Re: Derivatives of Exponential Functions

Hello, Kimmy!

What's stopping you from taking derivatives?

If $\displaystyle y\:=\:\frac{5}{2}\,\left(e^{\frac{x}{5}}\,+\,e^{-\frac{x}{5}}\right)$, prove that: $\displaystyle \: y''\:=\:\frac{1}{25}y$

We have: $\displaystyle \:y\;=\;\frac{5}{2}\,\left(e^{\frac{1}{5}x}\,+\,e^{-\frac{1}{5}x}\right)$

Then: $\displaystyle \:y'\;=\;\frac{5}{2}\,\left[\frac{1}{5}\cdot e^{\frac{1}{5}x}\,+\,\left(-\frac{1}{5}\right)\cdot e^{-\frac{1}{5}x}\right] \;=\;\frac{5}{2}\,\left[\frac{1}{5}\cdot e^{\frac{1}{5}x}\,-\,\frac{1}{5}\cdot e^{-\frac{1}{5}x}\right]$

Then: $\displaystyle \:y'' \;= \;\frac{5}{2}\,\left[\frac{1}{5}\cdot\frac{1}{5}\cdot e^{\frac{1}{5}x} \,-\,\frac{1}{5}\left(-\frac{1}{5}\right) e^{-\frac{1}{5}x}\right] \;=\;\frac{5}{2}\,\left[\frac{1}{25}\cdot e^{\frac{1}{5}x}\,+\,\frac{1}{25}\cdot e^{-\frac{1}{5}x}\right]$

And we have: $\displaystyle \:y'' \;= \;\frac{1}{25}\,\underbrace{\left[\frac{5}{2}\left(e^{\frac{1}{5}x}\,+\,e^{-\frac{1}{5}x}\right)\right]}$
. . Therefore: . . . $\displaystyle \;y''\;\;=\;\;\frac{1}{25}\cdot y$

 

[kimmy_koo51]

Ohh! Thank you that makes so much sence! I am home sick for a few days and trying to figure out like 3 days worth of Calculas lessons! It sucks