MHB Differentiation Help: Find 2nd Derivative

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To find the second derivative of the function L(λ) = λ^150 e^(-3λ), the first derivative is calculated as L'(λ) = 150λ^149 e^(-3λ) - 3λ^150 e^(-3λ), which simplifies to L'(λ) = 3λ^149 e^(-3λ)(50 - λ). The discussion emphasizes applying the product rule for derivatives, specifically stating that if L' = uvw, then the second derivative L'' can be found using the formula L'' = u'vw + uv'w + uvw'. Participants are encouraged to apply this method to compute L''. The thread focuses on the correct application of differentiation techniques for complex 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 ...
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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