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If $y=\frac{5}{2}\left(e^{\frac{x}{5}}+e^{-\frac{x}{5}}\right),$ prove that $y^{\prime \prime}=\frac{y}{25}$.

$\begin{aligned} \text { If } y &=\frac{5}{2}\left(e^{l}+e^{-\zeta}\right), \text { then } \\ y^{\prime} &=\frac{5}{2}\left(\frac{1}{5} e^{k}-\frac{1}{5} e^{-1}\right), \text {and } \\ y^{\prime \prime} &=\frac{5}{2}\left(\frac{1}{25} e^{\frac{1}{4}}+\frac{1}{25} e^{-4}\right) \\ &=\frac{1}{25}\left[\frac{5}{2}\left(e^{t}+e^{-t}\right)\right] \\ &=\frac{1}{25} y \end{aligned}$

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{'transcript': "section 3.6 problems, 77 were given a function and were asked to find its second derivative. Well, there's no shortcut to finding the second derivative without finding the first derivative. So let's differentiate this the differential oven exponents of an exponential function. It's just going to be e to the X squared according to the chain row. Now, what is the derivative of X squared? That's two x so plus five. So this is equal to what Two acts e to the X squared plus five. Now I need to differentiate that once more to get to my second derivative. I'm going to use the product rule on this first expression. So the derivative of two X is too at the derivative. Either the X squared is he to the X squared times two x plus. Um, sorry we back up. I was different, training both terms, So make sure you know what I did here, So I differentiated the two X that gives me a two times anything X squared now, plus two acts and now different shape. Either the X squared. That's either the X squared times two x plus, the derivative of five is simply zero. And so now if you look at what you've got, you've got to. So the second derivative is too e to the X squared. And then you're left with what? One plus, um, two x squared. And that is your final answer. So to eat of the X squared times one plus two x squared."}

Florida State University

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