Differentiation Help: Find 2nd Derivative

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SUMMARY

The discussion focuses on finding the second derivative of the function $L(\lambda) = \lambda^{150}e^{-3\lambda}$. The first derivative is calculated as $L’(\lambda) = 3\lambda^{149}e^{-3\lambda} (50-\lambda)$. Participants emphasize the application of the product rule for differentiation, specifically using the formula $L’’ = u’vw + uv’w + uvw’$ to derive the second derivative. This method ensures accurate differentiation of products of functions.

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I can't seem to find the second derivative
 

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$L(\lambda) = \lambda^{150}e^{-3\lambda}$

$L’(\lambda) = 150 \lambda^{149} e^{-3\lambda} - 3\lambda^{150} e^{-3\lambda}$

$L’(\lambda) = 3\lambda^{149}e^{-3\lambda} (50-\lambda)$

note ...

if $L’ = uvw$, where $u,v, \text{ and } w$ are all functions of $\lambda$, then ...

$L’’ = u’vw + uv’w + uvw’$

give it a go ...
 

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