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

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

## Derivative of xex Formula

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

• Product rule of derivatives.
• First principle of derivatives.

## Derivative of xex by Product Rule

The product rule is used to find the derivative of the product of functions. As xex is the product of x and ex, one can use the product rule to evaluate the derivative of xex. 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=ex.

Then by the above product rule, the derivative of xex is equal to

$(xe^x)’=x(e^x)’+e^x(x)’$

= xex + ex ⋅ 1

= xex + ex

= (x+1)ex

So the derivative of xex by the product rule of differentiation is equal to (x+1)ex.

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## Derivative of xex by First Principle

Let f(x) =xex. Then by the first principle of derivatives, that is, by the limit definition, the derivative of xex 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.

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

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

= xex ⋅ 1 + ex as the limit of (ex-1)/x is 1 when x0.

= xex+ex

So the derivative of xex by the first principle is equal to ex+xex.

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/x2.

## FAQs on Derivative of xex

Q1: What is the derivative of xex?

Answer: The derivative of xex is ex+xex.

Q2: What is the derivative of xe-x?

Answer: The derivative of xe-x is e-x-xe-x.

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