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 will see how to find the derivative of the natural logarithm of x squared. We will see proofs, graphical comparisons of the original function and its derivative, 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 \( y = \ln{\left(x^2\right)} \)

$$ y’ = \frac{2}{x} $$

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

The graph of the function

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

is

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

Comparing the graphs, we have

Using these graphs, you can see that the original function (f(x) = \ln(x^2) 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 \( y | y \neq 0 \)

## See also

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

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