# Derivative of xlogx: Proof by First Principle, Product Rule

The derivative of xlog x is equal to 1+log x. Note that x log x is the product of x with its logarithm. In this post, we will learn how to find the derivative of xlog x.

## Derivative of xlogx Formula

The formula for the derivative of xlogx is given as follows: d(xlogx)/dx = 1+logx. Here, the differentiation has been taken with respect to x. In the next two sections below, we will evaluate the derivative of xlogx using the product rule and the first principle of derivatives (i.e., the limit definition of derivatives).

## Derivative of xlogx by Product Rule

As xlogx is a product of two functions x and logx, we can find the derivative of xlogx by the product rule of differentiation. Here, log x is considered with base e.

We will use the derivatives of x and log x given below.

• d(x)/dx =1
• d(log x)/dx = 1/x.

Now, by the product rule of derivatives, the differentiation of xlogx is equal to

$\dfrac{d}{dx}(x\log x)$ $=x \dfrac{d}{dx}(\log x) + \log x \dfrac{d}{dx}(x)$

= $x \cdot \dfrac{1}{x}+\log x \cdot 1$

= $1+\log x$.

Hence, the derivative of xlog x by the product rule is 1+logx.

## Derivative of xlogx by First Principle

Let f(x)=xlogx. Then the derivative of xlogx from the first principle is given by the following limit formula:

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

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

Rearranging the numerator, the above limit

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

= $\lim\limits_{h \to 0} \dfrac{x\log \frac{x+h}{x}}{h}$ $+\lim\limits_{h \to 0} \dfrac{h \log(x+h)}{h}$ by the logarithm rule log a – log b = log a/b.

= $x\lim\limits_{h \to 0} \dfrac{\log \big(1+\frac{h}{x}\big)}{h/x} \times \dfrac{1}{x}$ $+\log(x+0)$

[Let t=h/x. Then t→0 as h→0]

= $x \times \dfrac{1}{x} \lim\limits_{t \to 0} \dfrac{\log \big(1+t\big)}{t}+\log x$

= $x \times \dfrac{1}{x} \times 1+\log x$ as the limit of log(1+x)/x is 1 when x→0.

= $1+\log x$

So the derivative of xlogx by the first principle is equal to 1+logx.

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

Derivative of xex: The derivative of xex is ex(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 xlogx

Q1: What is the derivative of logx?

Answer: The derivative of logx is 1/x.

Q2: What is the derivative of xlogx?

Answer: The derivative of xlogx is 1+logx.

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