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 first principle and by the product rule of derivatives.The first principle of derivatives says that the derivative of a function f(x) is given by the following limit formula:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

**$\cdots (\star)$**## Derivative of xe^{x} by First Principle

We will put f(x)=xe

^{x}in the above formula**$(\star)$**. By doing so the derivative of xe^{x}by the first principle, that is, by the limit definition is given as follows:$\dfrac{d}{dx}(xe^x)$ $=\lim\limits_{h \to 0} \dfrac{(x+h)e^{x+h}-xe^x}{h}$

$=\lim\limits_{h \to 0}$ $\dfrac{xe^{x+h}+he^{x+h}-xe^{x}}{h}$

$=\lim\limits_{h \to 0}$ $[\dfrac{xe^{x+h}-xe^{x}}{h}$ $+\dfrac{he^{x+h}}{h}]$

$=\lim\limits_{h \to 0}$ $\dfrac{xe^{x}e^{h}-xe^{x}}{h}$ $+\lim\limits_{h \to 0} e^{x+h}$

$=xe^{x} \lim\limits_{h \to 0}$ $\dfrac{e^{h}-1}{h}$ $+e^{x+0}$

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

= xe

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

^{x}.Hence, the derivative of xe

^{x}from the first principle is (1+x)e^{x}.## Derivative of xe^{x} by Product Rule

Let us first recall the 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 the rule below:

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

We have $f(x)=xe^{-x}$. So we put $g(x)=x$ and $h(x)=e^{x}$.

Hence, the derivative of $f(x)=xe^{x}$ by the product rule is given by

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

$=-xe^{x}+e^{x}\cdot 1$

= (1+x)e

^{x}.Therefore, the derivative of xe

^{x}by the product rule is equal to (1+x)e^{x}.