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f"(x) of 3e^(-x)^(2) I dont know how to find this. I am using it so that i can use the trapizodial rule
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is your question \[f(x)=3e^{-x^2}\]find \(f''(x)\) ?
What is the second derivative?
yes
can you find the first derivative using the chain rule?
I thought it would be 3e^(-x^2)*-6x
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if u=-x^2, then du/dx=-2x, so where did the -6x come from?
i did chain rule on it
\[u=-x^2\]\[f(x)=3e^{-x^2}=3e^u\]\[f'(x)=3e^u{du\over dx}\]
what is du/dx ?
Im really not sure.
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\[u=-x^2\implies{du\over dx}={d\over dx}(-x^2)=?\]
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