Question #e365f

Consider differentiating #e^x#:
#d/dxe^x=e^x#

It's a special function because its derivative is itself. Now, try #e^-x#:
#d/dxe^-x=-e^-x#

We see that the derivative is the same as itself again - except this time, it's negative. Why?

The answer has to do with the chain rule, which states that the derivative of a composite function (a function within a function) is the derivative of the inner function times the derivative of the function in general. This is best demonstrated with an example:
#d/dxe^(x^2)#

We have one function, #x^2#, nestled within another function, #e^x#. By the chain rule, the derivative is the derivative of #x^2# times the derivative of #e^(x^2)#:
#d/dxe^(x^2)=(x^2)'*(e^(x^2))'=color(red)(2x)color(blue)(e^(x^2))#
Because #d/dxx^2=color(red)(2x)# and #d/dxe^(x^2)=color(blue)(e^(x^2))#

The same logic holds for functions with negative exponents, like #e^-x#. In this case, we have #-x# nestled within #e^x#; the derivative is #(-x)'*(e^-x)'#:
#d/dxe^-x=-1*e^-x=-e^-x#

This makes intuitive sense too. We know that #e^-x# is constantly getting smaller, which means the slope is always negative. The derivative, #-e^-x# should reflect that, and it does: #-e^-x#, because of the negative sign put there by the chain rule, is always negative.

So for #f(x)=x^3e^-x#, you would use the product rule:
#d/dx(uv)=u'v+uv'#
In this case #u=x^3->u'=3x^2# and #v=e^-x->v'=-e^-x#
#d/dx(x^3e^-x)=(3x^2)(e^-x)+(x^3)(-e^-x)#
#=3x^2e^-x-x^3e^-x#