The derivative of *x* multiplied by the natural log of *x*, symboled as* x* ln(*x*), is the differentiation of a variable *x* multiplied by the logarithm of x to base *e* (Euler’s number). **The derivative of x ln(x) is equal to 1+ln(x).** This derivative can be found using the product rule of derivatives.

In this article, we’ll look at how to get the derivative of *x* ln(*x*). We will review some principles, definitions, formulas, graph comparisons of underived and derived *x* ln(*x*), proofs, derivation techniques, and examples.

## Proofs of the Derivative of Natural Logarithm, *x* ln*(x)*

Listed below are the proofs of the derivative of * \(x\ln{(x)}\)*. These proofs can also serve as the main methods of deriving this function.

### Proof of the derivative of *x ln(x)* using the Product Rule formula

In the derivative process of * \(x\ln{(x)}\)*, the product rule is used since the natural log of

*x*is being multiplied by another

*x*. Here, we have two multiplicands. They are two functions being multiplied together that we cannot algebraically simplify.

You can review the product rule formula by looking at this article: Product Rule of derivatives. You can also look at this article for the proof of the natural logarithm’s derivative using limits: Derivative of Natural log (ln(x)).

Let’s have the derivative of the function

$$ f(x) = x\ln{(x)}$$

We can figure out the two functions being multiplied. There is a natural logarithmic function and a monomial in this case. Setting the first multiplicand/term as *u*, we have

$$ u = x$$

and setting the second multiplicand/term as *v*, we have

$$ v = \ln{(x)}$$

Recall that the derivative product formula is

$$ \frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u$$

That is, the function *u times v* is derived by multiplying *u* by the derivative of *v* and then added to *v* multiplied by the derivative of *u*. Applying this formula to our given function, we have

$$ \frac{d}{dx} uv = (x) \frac{d}{dx} (\ln{(x)}) + (\ln{(x)}) \frac{d}{dx} (x)$$

Evaluating the derivative of *u* by using power rule and *v* by using the derivative of natural logarithm, we have

$$ \frac{d}{dx} uv = (x) \cdot \left(\frac{1}{x} \right) + (\ln{(x)}) \cdot (1)$$

Simplifying, we have

$$ \frac{d}{dx} uv = \frac{x}{x} + (\ln{(x)}) $$

$$ \frac{d}{dx} uv = 1 + (\ln{(x)}) $$

As a result, we arrive at the *\(x\ln{(x)}\)* derivative formula.

$$ \frac{d}{dx} x\ln{(x)} = 1 + (\ln{(x)}) $$

### Proof of the derivative of *x ln(x)* using implicit differentiation

You are advised to learn/review the derivatives of exponential functions and implicit differentiation for this proof.

Given that the equation

$$ y = x\ln{(x)}$$

Applying a logarithmic property

$$ y = \ln{(x)^x}$$

$$ y = \ln{\left(x^x\right)}$$

In general logarithmic form, it is

$$ \log_{e}{x^x} = y$$

And in exponential form, it is

$$ e^y = x^x$$

Implicitly deriving the exponential form in terms of *x*, we have

$$ e^y = x^x$$

$$ e^y \cdot \frac{d}{dx} = \frac{d}{dx} (x^x) $$

$$ e^y \cdot \frac{d}{dx} = \frac{d}{dx} (x^x + x^x \ln{(x)}) $$

Isolating \( \frac{dy}{dx} \), we have

$$ \frac{dy}{dx} = \frac{x^x + x^x \ln{(x)}}{e^y} $$

We recall that in the beginning, \( y = x\ln{(x)} \) or \( y = \ln{\left(x^x\right)} \). Substituting this to the *y* of our derivative, we have

$$ \frac{dy}{dx} = \frac{x^x + x^x \ln{(x)}}{e^{\left(\ln{\left(x^x\right)}\right)}} $$

Simplifying and applying a property of logarithm, we have

$$ \frac{dy}{dx} = \frac{x^x + x^x \ln{(x)}}{x^x} $$

$$ \frac{dy}{dx} = \frac{x^x}{x^x} + \frac{x^x \ln{(x)}}{x^x} $$

$$ \frac{dy}{dx} = 1 + \ln{(x)} $$

Evaluating, we now have the derivative of \( y = x\ln{(x)} \)

$$ y’ = 1 + \ln{(x)} $$

## Graph of *x ln(x)* vs. its derivative

In the instance of this function

$$ f(x) = x\ln{(x)}$$

the graph is illustrated as

And as we learned above, deriving \(f(x) = x\ln{(x)}\) will be

$$ f'(x) = 1 + \ln{(x)}$$

which is illustrated graphically as

Illustrating both graphs in one, we have

By examining the differences between these functions using these graphs, you can see that the original function \(f(x) = x\ln{(x)}\) has a domain of

\( (0,\infty) \) or \( x | x > 0 \)

and lies within the range of

\( \left[-\frac{1}{e}, \infty \right) \) or \( y | y \geq -\frac{1}{e} \)

whereas the derivative \(f'(x) = 1 + \ln{(x)}\) has a domain of

\( (0,\infty) \) or \( x | x > 0 \)

which lies within the range of

\( (-\infty,\infty) \) or *all real numbers*

## Examples

The example below show how to derive a variable multiplied by the natural logarithmic of the same variable using the product rule formula.

### EXAMPLE 1

**Derive:** \(f(\beta) = \beta\ln{(\beta)}\)

**Solution:** Using the product rule formula, we have

**Step 1:** Analyze if \( \beta\ln{(\beta)} \) is a function of the same variable \(\beta\). In this problem, it is. Hence, proceed to step 2.

**Step 2: **Set the first multiplicand/term of \(f(\beta) \) as *u* and the second as *v*.

$$ u = \beta$$

$$ v = \ln{(\beta)} $$

**Step 3:** Apply the product rule formula.

$$ \frac{d}{d\beta} uv = u \frac{d}{d\beta} v + v \frac{d}{d\beta} u$$

$$ \frac{d}{d\beta} uv = (\beta) \frac{d}{d\beta} (\ln{(\beta)}) + (\ln{(\beta)}) \frac{d}{d\beta} (\beta)$$

**Step 4:** Evaluate the derivatives of *u* and *v*.

$$ \frac{d}{d\beta} uv = (\beta) \cdot \left(\frac{1}{\beta}\right) + (\ln{(\beta)}) \cdot (1)$$

**Step 5:** Simplify and declare the final answer.

$$ \frac{d}{d\beta} uv = \left(\frac{\beta}{\beta}\right) + (\ln{(\beta)})$$

**The final answer is:**

$$ \frac{d}{d\beta} \beta\ln{(\beta)} = 1 + \ln{(\beta)} $$

## See also

Interested in learning more about the derivatives of logarithmic functions? Take a look at these pages:

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