# What is the slope of #f(x)=xe^(x-x^2) # at #x=-1#?

The slope of the function at x=-1 is

Applying product rule,

We get,

Simplifying it,

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To find the slope of the function ( f(x) = xe^{x-x^2} ) at ( x = -1 ), we first need to find its derivative, then evaluate it at ( x = -1 ).

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

Now, evaluate ( f'(-1) ):

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

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

( f'(-1) = e^{-2} - 3e^{-2} )

( f'(-1) = (1 - 3)e^{-2} )

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

Therefore, the slope of ( f(x) = xe^{x-x^2} ) at ( x = -1 ) is ( -2e^{-2} ).

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