Question

Question #e365f

Answer 1

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`