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 First Principle

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

$(xe^{-x})’$ = lim_{h→0} $\dfrac{(x+h)e^{-(x+h)}-xe^{-x}}{h}$

= lim_{h→0} $\dfrac{xe^{-(x+h)}+he^{-(x+h)}-xe^{-x}}{h}$

= lim_{h→0} $\big(\dfrac{xe^{-(x+h)}-xe^{-x}}{h}$ $+\dfrac{he^{-(x+h)}}{h} \big)$

= lim_{h→0} $\dfrac{xe^{-x}e^{-h}-xe^{-x}}{h}$ + lim_{h→0} $e^{-(x+h)}$

= xe^{-x} lim_{h→0} $\dfrac{e^{-h}-1}{h}$ $+e^{-(x+0)}$

= xe^{-x} lim_{h→0} $\dfrac{e^{-h}-1}{-h} \times (-1)$ +e^{-x}

Let z=-h. Then z →0 as h →0.

= -xe^{-x} lim_{h→0} $\dfrac{e^{z}-1}{z}$ $+e^{-x}$

= -xe^{-x} ⋅ 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}.

## Derivative of xe^{-x} by Product Rule

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

$\dfrac{d}{dx}(f(x))$ $=g(x) \dfrac{d}{dx}(h(x))+h(x) \dfrac{d}{dx}(g(x))$

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

Take g(x)=x and h(x)=e^{-x} in the product rule.

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

$\dfrac{d}{dx}(xe^{-x})$ $=x \dfrac{d}{dx}(e^{-x})+e^{-x} \dfrac{d}{dx}(x)$

= – xe^{-x} +e^{-x} ⋅ 1

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

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

**Also Read:**

Derivative of root x + 1 by root x

## FAQs

### Q1: What is the derivative of xe^{-x}?

Answer: The derivative of xe^{-x} is equal to (1-x)e^{-x}.