# Derivative of xe^-x by Product Rule, First Principle

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.