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.

Read These: Derivative of xx

Derivative of e-x | Derivative of sin3x

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.

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/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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