# Derivative of ln(x^2) with Proofs and Graphs

The natural logarithm of x squared, also denoted as ln(x2), is the logarithm of x2 to base e (euler’s number)The derivative of the natural logarithm of x2 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.

See proofs

##### CALCULUS

Relevant for

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

See proofs

## 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.

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

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

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

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

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

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