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

1 Answer
Feb 16, 2016

The slope of the function at x=-1 is -2/e^2

Explanation:

In order to find the slope of a function,we have to firstly determine its derivative.
so,
d/dx(xe^(x-x^2))

Applying product rule,

(f.g)'=f'.g+g.f'

so, f=x,\g=e^{x-x^2}

\frac{d}{dx}(x)e^{x-x^2}+\frac{d}{dx}(e^{x-x^2})x

We know...
d/dx(x)=1

and,
\frac{d}{dx}(e^{x-x^2})
Applying chain rule, \frac{df(u\)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}

let x-x^2=u

\frac{d}{du}(e^u\)\frac{d}{dx}\(x-x^2)

We get,

(e^u)(1-2x)

Substituting back x-x^2=u,

=e^{x-x^2}(1-2x)

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

Simplifying it,

e^{x-x^2}(-2x^2+x+1)

Finally when x=-1
e^{(-1)-(-1)^2}[-2(-1)^2+(-1)+1] = e^-2 . (-2)= -2/e^2