The natural logarithm of x squared, also denoted as ln(x^{2}), is the logarithm of *x*^{2} to base *e (euler’s number)*. **The derivative of the natural logarithm of x ^{2} is equal to two over x, 2/x.** We can prove this derivative using the chain rule or implicit differentiation.

In this article, we’ll look at how to find the derivative of the natural logarithm of x squared. We will go over some fundamentals, definitions, formulas, graph comparisons of the original function and its derivative, proofs, derivation techniques, and some examples.

##### CALCULUS

**Relevant for**…

Learning how to find the derivative of the natural logarithm of x squared.

##### CALCULUS

**Relevant for**…

Learning how to find the derivative of the natural logarithm of x squared.

## Proofs of the Derivative of Natural Logarithm of *x^2*

Listed below are the proofs of the derivative of \(\ln{\left(x^2\right)}\). These proofs can also serve as the main methods of deriving this function.

### Proof of the derivative of \(\ln{\left(x^2\right)}\) using the Chain Rule Formula

Given that this is a composed function, the chain rule formula can be used to prove the derivative formula for the natural log of \(x^2\). In the composite function \(\ln{\left(x^2\right)}\), the natural logarithmic function will be the outer function *f(u)*, while the \(x^2\) will be the inner function *g(x)*.

As a prerequisite for this topic, please review the chain rule formula by looking at this article: Chain Rule of derivatives. You may also check out this article for the proof of the derivative of natural logarithm using the *first principle*: Derivative of Natural log (ln(x)).

Let’s have the derivative of the function

$$ F(x) = \ln{\left(x^2\right)}$$

We can determine the two functions that comprise *F(x)*. In this instance, there is a natural logarithmic function and a monomial. We may configure the outer function as follows:

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

where

$$ u = x^2$$

Setting the monomial \(x^2\) as the inner function of *f(u)* by denoting it as *g(x)*, we have

$$ f(u) = f(g(x))$$

$$ g(x) = x^2$$

$$ u = g(x)$$

Deriving the outer function *f(u)* using the derivative of natural log in terms of *u*, we have

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

$$ f'(u) = \frac{1}{u}$$

Deriving the inner function *g(x)* using power rule since it is a monomial, we have

$$ g(x) = x^2$$

$$ g'(x) = 2x$$

Algebraically multiplying the derivative of outer function $latex f'(u)$ by the derivative of inner function $latex g'(x)$, we have

$$ \frac{dy}{dx} = f'(u) \cdot g'(x)$$

$$ \frac{dy}{dx} = \left(\frac{1}{u} \right) \cdot (2x)$$

Substituting *u* into *f'(u)* and simplifying, we have

$$ \frac{dy}{dx} = \left(\frac{1}{\left(x^2\right)} \right) \cdot (2x)$$

$$ \frac{dy}{dx} = \frac{2x}{x^2}$$

$$ \frac{dy}{dx} = \frac{2}{x}$$

As a result, we arrive at the \(\ln{\left(x^2\right)}\) derivative formula.

$$ \frac{d}{dx} \ln{\left(x^2\right)} = \frac{2}{x}$$

### Proof of the derivative of \(\ln{\left(x^2\right)}\) using implicit differentiation

In this proof, you are hereby recommended to learn/review the derivatives of exponential functions and implicit differentiation.

Suppose we have the equation

$$ y = \ln{\left(x^2\right)}$$

In general logarithmic form, it is

$$ \log_{e}{x^2} = y$$

And in exponential form, it is

$$ e^y = x^2$$

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

$$ e^y = x^2$$

$$ \frac{d}{dx} (e^y) = \frac{d}{dx} (x^2) $$

$$ e^y \cdot \frac{dy}{dx} = 2x $$

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

$$ \frac{dy}{dx} = \frac{2x}{e^y} $$

We recall that in the beginning, \( y = \ln{\left(x^2\right)} \). Substituting this to the *y* of our derivative, we have

$$ \frac{dy}{dx} = \frac{2x}{e^{\left(\ln{\left(x^2\right)}\right)}} $$

Simplifying and applying a property of logarithm, we have

$$ \frac{dy}{dx} = \frac{2x}{x^2} $$

$$ \frac{dy}{dx} = \frac{2}{x} $$

Evaluating, we now have the derivative of \( y = \ln{\left(x^2\right)} \)

$$ y’ = \frac{2}{x} $$

## Other Methods to derive the Natural Logarithmic of *x^2*

Aside from the two proofs given above which also serve as the main methods of deriving *\(\ln{\left(x^2\right)}\)*, an additional method can also be used to derive this same function.

### Derivative of the natural logarithmic of *x^2* by using the derivative of general logarithmic function.

**Step 1:** Express the function as \(f(x) = \log_{e}{x^2}\) instead of \(\ln{\left(x^2\right)}\)

** Step 2:** Take note that the derivative formula for a general logarithmic function is

$$ \frac{d}{dx} (\log_{b}{u}) = \frac{1}{u\ln{(b)}} \cdot \frac{d}{dx} (u)$$

where *b* is any positive real number except 0 and 1 and *u *is any function other than *x* which is the argument of the logarithmic function.

** Step 3:** Get the derivative of \(f(x) = \log_{e}{x^2}\) by applying the formula in step 2.

In this case, our *u* is \(x^2\). Hence,

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{1}{x^2\ln{(e)}} \cdot \frac{d}{dx} (x^2)$$

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{1}{x^2\ln{(e)}} \cdot (2x)$$

** Step 4:** Algebraically evaluate and simplify.

Evaluating *ln(e)*, it is just equal to one. Hence,

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{1}{x^2 \cdot (1)} \cdot (2x)$$

Simplifying, we have

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{1}{x^2} \cdot (2x)$$

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{2x}{x^2} $$

$$ \frac{d}{dx} \left(\log_{e}{x^2}\right) = \frac{2}{x} $$

** Step 5:** Final Answer.

$$ \frac{d}{dx} \left(\ln{\left(x^2\right)}\right) = \frac{2}{x}$$

## Graph of ln(x^2) vs. its derivative

Given the function

$$ f(x) = \ln{\left(x^2\right)}$$

the graph is illustrated as

And as we know by now, by deriving \(f(x) = \ln{\left(x^2\right)}\), we get

$$ f'(x) = \frac{2}{x}$$

which is illustrated graphically as

Illustrating both graphs in one, we have

Analyzing the differences of these functions through these graphs, you can observe that the original function \(f(x) = \ln{\left(x^2\right)}\) has a domain of

\( (-\infty,0) \cup (0,\infty) \) or \( x | x \neq 0 \)

and exists within the range of

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

whereas the derivative \(f'(x) = \frac{2}{x}\) has a domain of

\( (-\infty,0) \cup (0,\infty) \) or \( x | x \neq 0 \)

and exists within the range of

\( (-\infty,0) \cup (0,\infty) \) or \( y | y \neq 0 \)

## Examples

Below are some examples of deriving a natural logarithmic of a variable raised to two through the use of chain rule or general logarithmic derivative.

### EXAMPLE 1

**Derive:** \(F(\beta) = \ln{\left(\beta^2\right)}\)

**Solution:** Using chain rule, we have

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

**Step 2: **Isolate the two functions composed in \( F(\beta) \). The outer function is

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

and the inner function is

$$ u = g(\beta) = \beta^2 $$

**Step 3:** Derive both outer and inner functions separately using their respective base derivative formula.

$$ f'(u) = \frac{d}{du} \ln{(u)} = \frac{1}{u} $$

$$ g'(\beta) = \frac{d}{d\beta} \beta^2 = 2\beta $$

**Step 4:** Multiply the derivative of *f(u)* and *g(\beta)*.

$$ F'(\beta) = \left( \frac{1}{u} \right) \cdot (2\beta) $$

**Step 5:** Substitute *u* into the derivative of *f(u)*.

$$ F'(\beta) = \left( \frac{1}{(\beta^2)} \right) \cdot (2\beta) $$

**Step 6:** Simplify and declare the final answer.

$$ F'(\beta) = \frac{2\beta}{(\beta^2)} $$

**The final answer is:**

$$ \frac{d}{d\beta} \ln{\left(\beta^2\right)} = \frac{2}{(\beta)} $$

### EXAMPLE 2

**Derive:** \(f(\lambda) = \ln{\left(\lambda^2\right)}\)

**Solution:** Using the general logarithmic derivative formula, we have

**Step 1:** Express the function as \(f(\lambda) = \log_{e}{\lambda^2}\) instead of \(\ln{(\lambda^2)}\)

**Step 2:** Set the logarithmic argument as *u*.

$$ u = x^2 $$

**Step 3:** Recalling the derivative formula for a general logarithmic function, we have

$$ \frac{d}{dx} (\log_{b}{u}) = \frac{1}{u\ln{(b)}} \cdot \frac{d}{dx} (u)$$

**Step 4:** Derive \(f(\lambda) = \log_{e}{\lambda}^2\) by applying the formula in step 2.

$$ \frac{d}{d\lambda} \left(\log_{e}{\lambda^2}\right) = \frac{1}{u\ln{(e)}} \cdot \frac{d}{d\lambda} (u)$$

$$ \frac{d}{d\lambda} \left(\log_{e}{\lambda^2}\right) = \frac{1}{\lambda^2\ln{(e)}} \cdot \frac{d}{dx} (\lambda^2)$$

$$ \frac{d}{d\lambda} \left(\log_{e}{\lambda^2}\right) = \frac{1}{\lambda^2 \cdot (1)} \cdot (2\lambda)$$

$$ \frac{d}{d\lambda} \left(\log_{e}{\lambda^2}\right) = \frac{2\lambda}{\lambda^2}$$

$$ \frac{d}{d\lambda} \left(\log_{e}{\lambda^2}\right) = \frac{2}{\lambda}$$

$$ \frac{d}{d\lambda} \left(\ln{(\lambda^2)}\right) = \frac{2}{\lambda}$$

## See also

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

### Learn mathematics with our additional resources in different topics

**LEARN MORE**