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Mathematics 12 Online
OpenStudy (anonymous):

How would you take the derivative of: f(x) = xe^(-kx)? Please explain how to take derivatives of the variable 'e'.

OpenStudy (anonymous):

another product rule problem

OpenStudy (anonymous):

the derivative of \(e^x\) is \(e^x\) so by the chain rule the derivative of \(e^{-kx}\) is \(-ke^{-kx}\) then apply the product rule

OpenStudy (anonymous):

you must understand one thing that e is not a variable its a constant e^x is a variable.............................

OpenStudy (anonymous):

as before it is \[(fg)'=f'g+g'f\] with \[f(x)=x,f'(x)=1,g(x)=e^{-kx},g'(x)=-ke^{-kx}\]

OpenStudy (anonymous):

actually \(x\) is a variable, \(e^x\) is a function, although it too varies according to \(x\)

OpenStudy (anonymous):

So whenever you have something like: e^(ax), you would use the chain rule etc?

OpenStudy (anonymous):

although e^x is a function but still it varies with x hence its a variable.......

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