# Thread: Deriviative of function # 8

1. ## Deriviative of function # 8

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

Ok can someone just check with me.. I'm pretty close to the answer but i don't know what I did wrong ..

the final answer of y''=y/25.. I have

[e^(x/5)+e^(-x/5)]/[e^(x/5)/25+e^(-x/5)/25]

Don't know what happened to the 5/2 in y and.. the denominator is almost there but not quite. ^.^

2. Presumably this is what you got (please learn latex):
$\displaystyle y = \frac 5 2 (e^{x/5} + e^{-x/5})$

Diffing $\displaystyle e^{ax}$ gives $\displaystyle ae^{ax}$ and doing it twice gives $\displaystyle a^2e^{ax}$.

Diffing $\displaystyle e^{-ax}$ gives $\displaystyle -ae^{-ax}$ and doing it twice gives $\displaystyle (-a)^2e^{-ax} = a^2e^{ax}$.

Adding them together gives $\displaystyle \frac {d^2} {dx^2} (e^{x/5} + e^{-x/5}) = \frac 1 {25} (e^{x/5} + e^{-x/5})$.

Multiplying the lot by $\displaystyle \frac 5 2$ gives:
$\displaystyle \frac {d^2y} {dx^2} = \frac 5 2 \frac 1 {25} (e^{x/5} + e^{-x/5}) = \frac 1 {25} y$