The derivative of xe^-x is equal to (1-x)e^-x. In this post, we will find the derivative of xe^-x by the product rule and the first principle of derivatives.

## Derivative of xe^-x by Product Rule

Product rule of derivatives: If f(x) = g(x) h(x) is a product of two functions, then the derivative of f(x) is given by

`d/dx[f(x)]` `=g(x) d/dx[h(x)]+h(x) d/dx[g(x)]`

We have `f(x)=xe^{-x}`.

So we can take `g(x)=x` and `h(x)=e^{-x}`.

Thus, the derivative of `f(x)=xe^{-x}` by the product rule is equal to

`d/dx[xe^{-x}]` `=x d/dx[e^{-x}]+e^{-x} d/dx[x]`

`=-xe^{-x}+e^{-x}cdot 1`

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

Hence, the derivative of xe^-x by the product rule is (1-x)e^-x.

## Derivative of xe^-x by First Principle

The derivative of `f(x)=xe^{-x}`by the first principle is given by

`(xe^{-x})’` `=lim_{h to 0} frac{(x+h)e^{-(x+h)}-xe^{-x}}{h}`

`=lim_{h to 0}` `frac{xe^{-(x+h)}+he^{-(x+h)}-xe^{-x}}{h}`

`=lim_{h to 0}` `[frac{xe^{-(x+h)}-xe^{-x}}{h}` `+frac{he^{-(x+h)}}{h}]`

`=lim_{h to 0}` `frac{xe^{-x}e^{-h}-xe^{-x}}{h}` `+lim_{h to 0} e^{-(x+h)}`

`=xe^{-x} lim_{h to 0}` `frac{e^{-h}-1}{h}` `+e^{-(x+0)}`

`=xe^{-x} lim_{h to 0}` `frac{e^{-h}-1}{-h} times (-1)` `+e^{-x}`

Let `z=-h`. Then z tends to 0 as h tends to 0.

`=-xe^{-x} lim_{z to 0}` `frac{e^{z}-1}{z}` `+e^{-x}`

`=-xe^{-x} cdot 1 +e^{-x}` as we know that the limit of (e^z-1)/z tends to 1 when z tends to 0.

`=-xe^{-x} +e^{-x}` `=(1-x)e^{-x}`.

Hence, the derivative of xe^-x from the first principle is equal to (1-x)e^-x.