Derivative of x ln(x) with Proofs and Graphs

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.

CALCULUS
Derivative of x ln(x)

Relevant for

Learning how to find the derivative of x ln(x).

See proof

CALCULUS
Derivative of x ln(x)

Relevant for

Learning how to find the derivative of x ln(x).

See proof

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

graph-of-fx-xlnx

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

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

which is illustrated graphically as

graph-of-the derivative of-xlnx

Illustrating both graphs in one, we have

graph-of xlnx and its derivative

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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