derivative of y = f(x) = xe^(-x)

[Ashley5]

I started with

f'(x) = xe^(-x) (-1) + e^(-x) (1)

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

How do you finish it? Thank you

 

[o_O]

Not much else you can do besides factor out the e^-x to make it look "nicer" I guess.

 

[Ashley5]

so it would be
e^-x(-x+1)

Thank you!

 

[Ashley5]

how do you find the equation of the tangent line at the point (1,e^-1)

Since I did take the derivative of y=xe^-x

f'(x)= e^-x (-x+1)

If I put it in slope form y-y1=m(x-x1)
It just doesn't make since. I think I did the derivative wrong. Can someone help, please. Thank you!

 

[o_O]

Nothing wrong the derivative. If you're given the point (1, e^-1), then just plug x = 1 into f'(x) to get your slope. Then, use your "slope form" to get your tangent line.