F(x)=2xe^(-x) what is derivative?

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SUMMARY

The derivative of the function f(x) = 2xe^(-x) is f'(x) = (2 - 2x)e^(-x). The discussion emphasizes the application of the product rule for differentiation and simplifies the derivative effectively. The critical point identified is x = 1, which is confirmed to be a maximum value of the function. This conclusion is reached by setting the derivative to zero and analyzing the behavior of the function at this point.

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f(x)=2xe^(-x)

what is derivative?

f(x)=(2x)/(e^x)
f'(x)={{2e^x}-{2x^2e^{x-1}}}/(e^x)^2

can I simplify this?
 
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Recall that:
\frac {d e^x} {dx} = e^x
then apply the chain rule.
 
Last edited:
Surely easier to do by product rule!

f'(x)=2e^-^x - 2xe^-^x

or even:

f'(x)=(2 - 2x)e^-^x
 
Last edited:
gee thanks!
The questions acually asks to find each critical point of f and whether f(x) is a relative maximum, relative minimum, or neither. I set f'(x) to zero and solved. I got x=1.

I found it to be a maximum value. Is this correct?
 

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