Can anyone help me find the derivative of 3xe^-kx -thank you
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use chain rule, $\displaystyle u=3x$ and $\displaystyle v=e^{-kx}$
don't you have to use a combination of the product rule and chain rule?
Originally Posted by consevans Can anyone help me find the derivative of 3xe^-kx If $\displaystyle y=f(x)e^{g(x)}$ then $\displaystyle y'=f'(x)e^{g(x)}+f(x)g'(x)e^{g(x)}$
Originally Posted by BAdhi use chain rule, $\displaystyle u=3x$ and $\displaystyle v=e^{-kx}$ I think you mean "product rule".
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