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Numerade Educator

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Problem 50 Hard Difficulty

Find $ y' $ and $ y". $
$ y = e^{e^x} $

Answer

$\begin{aligned} y^{\prime} &=e^{x} \cdot e^{e^{x}} \\ y^{\prime \prime} &=e^{e^{x}+x}\left(e^{x}+1\right) \end{aligned}$

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Video Transcript

okay. Our function here is why equals E to the E to the X, and we're going to use the chain rule to find the derivative The derivative of E to the G of X will be e to the G of X times J prime of X. So for white crime, we have the derivative of E to the each of the X, which is each of the each of the X times, the derivative of each of the X, which is he to the X? Okay, we could use exponents properties where if you have a to the end times a to the M, you get a to the end plus M and we could add the exponents on E. We could add the E to the X and the X, and another form of our answer would be e to the each of the X plus X power. And I think we'll use that as we find the second derivative. So second derivative again, the derivative of each of the G of X is e to the g of x times G prime of X. So the derivative will be e to the E to the X plus X times, the derivative of each of the X plus X and the derivative of that is each of the X plus one.