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Mathematics 20 Online
OpenStudy (anonymous):

how to you find the derivative of e^(bx^2)?? @MIT 18.02 Multiva…

OpenStudy (anonymous):

with respect to x right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[e^{bx^2}\times 2bx\] then by the chain rule

OpenStudy (anonymous):

thx

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