Question #e365f

1 Answer
Jun 2, 2016

Same as you would if the exponent were positive - with a slight adjustment for the chain rule.

Explanation:

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