The derivative of xe^{x} is equal to (x+1)e^{x}. Note that xe^{x} is the product of two functions x and e^{x}. In this post, we will find the derivative of xe^{x} by the first principle and the product rule of derivatives.

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## Derivative of xe^{x} Formula

The formula for the derivative of xe^{x} is given by d(xe^{x})/dx = e^{x}+xe^{x}. Here the differentiation is taken with respect to the variable x. In the next sections, we will find the derivative of the product xe^{x} using the following methods:

- Product rule of derivatives.
- First principle of derivatives.

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

The product rule is used to find the derivative of the product of functions. As xe^{x} is the product of x and e^{x}, one can use the product rule to evaluate the derivative of xe^{x}. By the product rule, the derivative of uv is given by the formula:

(uv)$’$ = u v$’$ + v u$’$.

Here the prime denotes the derivative with respect to x.

Put u=x and v=e^{x}.

Then by the above product rule, the derivative of xe^{x} is equal to

$(xe^x)’=x(e^x)’+e^x(x)’$

= xe^{x} + e^{x} ⋅ 1

= xe^{x} + e^{x}

= (x+1)e^{x}

So the derivative of xe^{x} by the product rule of differentiation is equal to (x+1)e^{x}.

**Read These:** Derivative of x^{x}

Derivative of e^{-x} | Derivative of sin3x

## Derivative of xe^{x} by First Principle

Let f(x) =xe^{x}. Then by the first principle of derivatives, that is, by the limit definition, the derivative of xe^{x} is equal to

$(xe^x)’$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$

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

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

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

= $\lim\limits_{h \to 0}\dfrac{xe^xe^h-xe^x}{h}$ $+\lim\limits_{h \to 0}\dfrac{he^{x+h}}{h}$ by the sum rule of limits.

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

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

= xe^{x} ⋅ 1 + e^{x} as the limit of (e^{x}-1)/x is 1 when x**→**0.

= xe^{x}+e^{x}

So the derivative of xe^{x} by the first principle is equal to e^{x}+xe^{x}.

**Also Read:**

**Derivative of xcosx**: The derivative of xcosx is cosx – xsinx.

**Derivative of log3x**: The derivative of log3x is 1/x.

**Derivative of 1**: The derivative of 1 is zero.

**Derivative of 1/x**: The derivative of 1/x is -1/x^{2}.

## FAQs on Derivative of xe^{x}

**Q1: What is the derivative of xe**

^{x}?Answer: The derivative of xe^{x} is e^{x}+xe^{x}.

**Q2: What is the derivative of xe**

^{-x}?Answer: The derivative of xe^{-x} is e^{-x}-xe^{-x}.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.