\\[
\n\\frac{d}{dx}\\bigl(\\ln^2(x)\\bigr)=\\bigl(\\ln^2(x)\\bigr)’=\\frac{2 \\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
This is the main formula in this topic, and the later transformations are based on it. It applies to a function defined only for \\( x>0 \\), since the natural logarithm exists only for positive values of the argument.<\/p>\n
![\"Graph](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/04\/derivative-of-the-natural-logarithm-squared1.jpg\")
<\/p>\n
Now let us look at the graphs. The function \\( y=\\ln^2(x) \\) never takes negative values because the logarithm is squared. At the same time, its derivative \\( y’=\\frac{2 \\cdot \\ln(x)}{x} \\) can be either negative or positive. On the interval \\( 0<x<1 \\), we have \\( \\ln(x)<0 \\), so the derivative is negative. At the point \\( x=1 \\), it is equal to zero. Also, since \\( \\ln(1)=0 \\), we get \\( \\ln^2(1)=0 \\). This means that the function reaches its minimum value exactly at \\( x=1 \\). For \\( x>1 \\), the derivative becomes positive. So, the graph clearly shows that the function decreases on the interval \\( (0,1) \\) and then increases on \\( (1,+\\infty) \\).<\/p>\n
Step by Step: How to Derive the Formula<\/h2>\n
Now let us move on to the proof. The best place to start is with a careful reading of the function itself:<\/p>\n
\\[
\ny=\\ln^2(x)=\\bigl(\\ln(x)\\bigr)^2.
\n\\]<\/p>\n
This notation means that we first calculate the natural logarithm of \\( x \\), and then square the result. So, we are dealing with a composite function, and the natural method here is the chain rule. This is exactly the rule we use when one function is nested inside another.<\/p>\n
To write the proof clearly and step by step, let us introduce a new variable:<\/p>\n
\\[
\nu=\\ln(x).
\n\\]<\/p>\n
Then the original function takes a simpler form:<\/p>\n
\\[
\ny=u^2.
\n\\]<\/p>\n
Now let us find the derivative of the outer function. If \\( y=u^2 \\), then<\/p>\n
\\[
\n\\frac{dy}{du}=2 \\cdot u.
\n\\]<\/p>\n
After that, we find the derivative of the inner function. Since \\( u=\\ln(x) \\), we have<\/p>\n
\\[
\n\\frac{du}{dx}=\\frac{1}{x}.
\n\\]<\/p>\n
Next, we apply the chain rule:<\/p>\n
\\[
\n\\frac{dy}{dx}=\\frac{dy}{du}\\cdot\\frac{du}{dx}.
\n\\]<\/p>\n
Now substitute the derivatives we found:<\/p>\n
\\[
\n\\frac{dy}{dx}=2\\cdot u\\cdot\\frac{1}{x}.
\n\\]<\/p>\n
Then we return to the original variable \\( x \\), which means replacing \\( u \\) with \\( \\ln(x) \\):<\/p>\n
\\[
\n\\frac{dy}{dx}=2\\cdot\\ln(x)\\cdot\\frac{1}{x}.
\n\\]<\/p>\n
After simplification, we get<\/p>\n
\\[
\n\\frac{dy}{dx}=\\frac{2\\cdot\\ln(x)}{x}.
\n\\]<\/p>\n
So,<\/p>\n
\\[
\n\\bigl(\\ln^2(x)\\bigr)’=\\frac{2\\cdot\\ln(x)}{x}.
\n\\]<\/p>\n
Thus, the formula for the derivative of the natural logarithm squared follows directly from the chain rule. First, we differentiate the square, and then we multiply by the derivative of the inner function \\( \\ln(x) \\). That exact order gives the correct expression for the derivative.<\/p>\n
Derivative of the Natural Logarithm Squared: Practical Applications<\/h2>\n
Theory becomes much easier to understand when you can immediately see how it works in specific expressions. That is why below we will look at several problems in which the derivative of the natural logarithm squared is used together with other rules of differentiation. This helps us see more clearly where the known formula is applied directly, and where we also need to take into account a sum, a product, or another composite function.<\/p>\n
Example 1. Find the derivative of the function \\( y=x \\cdot \\ln^2(x) \\)<\/h3>\n
Here we have a product of two functions: \\( x \\) and \\( \\ln^2(x) \\). So, we apply the product rule:<\/p>\n
\\[
\ny’=(x)’\\cdot \\ln^2(x)+x\\cdot \\bigl(\\ln^2(x)\\bigr)’.
\n\\]<\/p>\n
The first term is found immediately because \\( (x)’=1 \\). So we get<\/p>\n
\\[
\ny’=\\ln^2(x)+x\\cdot \\bigl(\\ln^2(x)\\bigr)’.
\n\\]<\/p>\n
Now let us focus on the part \\( \\ln^2(x) \\). This is exactly where we use the formula we already know:<\/p>\n
\\[
\n\\bigl(\\ln^2(x)\\bigr)’=\\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
Substitute it into the previous expression:<\/p>\n
\\[
\ny’=\\ln^2(x)+x\\cdot \\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
Now we simplify \\( x \\) in the second term. So we get<\/p>\n
\\[
\ny’=\\ln^2(x)+2\\cdot \\ln(x).
\n\\]<\/p>\n
Example 2. Find the derivative of the function \\( y=\\dfrac{\\ln^2(x)}{x} \\)<\/h3>\n
Here we have a quotient, so we apply the quotient rule. Let<\/p>\n
\\[
\nu=\\ln^2(x),\\quad v=x.
\n\\]<\/p>\n
Then<\/p>\n
\\[
\ny’=\\frac{u’\\cdot v-u\\cdot v’}{v^2}.
\n\\]<\/p>\n
First, let us find the derivative of the denominator:<\/p>\n
\\[
\nv’=(x)’=1.
\n\\]<\/p>\n
Now we move to the numerator. We need to find \\( u’=\\bigl(\\ln^2(x)\\bigr)’ \\). Once again, we use the formula we already know:<\/p>\n
\\[
\nu’=\\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
After that, we substitute everything into the quotient rule:<\/p>\n
\\[
\ny’=\\frac{\\frac{2\\cdot \\ln(x)}{x}\\cdot x-\\ln^2(x)\\cdot 1}{x^2}.
\n\\]<\/p>\n
In the first term of the numerator, we simplify \\( x \\). So we obtain<\/p>\n
\\[
\ny’=\\frac{2\\cdot \\ln(x)-\\ln^2(x)}{x^2}.
\n\\]<\/p>\n
Example 3. Find the derivative of the function \\( y=\\ln^2(3\\cdot x+1) \\)<\/h3>\n
This is a composite function in which the outer part is the square of the natural logarithm, and the inner part is the expression \\( 3\\cdot x+1 \\). So, the chain rule is used here.<\/p>\n
Let us denote<\/p>\n
\\[
\nu=3\\cdot x+1.
\n\\]<\/p>\n
Then<\/p>\n
\\[
\ny=\\ln^2(u).
\n\\]<\/p>\n
Now we use the formula we already learned for the derivative of the square of the natural logarithm, but here the variable is \\( u \\) instead of \\( x \\):<\/p>\n
\\[
\n\\frac{d}{du}\\ln^2(u)=\\frac{2\\cdot \\ln(u)}{u}.
\n\\]<\/p>\n
Since \\( u \\) depends on \\( x \\), we also need to multiply by the derivative of the inner expression:<\/p>\n
\\[
\n\\frac{du}{dx}=(3\\cdot x+1)’=3.
\n\\]<\/p>\n
Then, by the chain rule, we have<\/p>\n
\\[
\ny’=\\frac{2\\cdot \\ln(u)}{u}\\cdot \\frac{du}{dx}.
\n\\]<\/p>\n
Substitute \\( u=3\\cdot x+1 \\) and \\( \\frac{du}{dx}=3 \\):<\/p>\n
\\[
\ny’=\\frac{2\\cdot \\ln(3\\cdot x+1)}{3\\cdot x+1}\\cdot 3.
\n\\]<\/p>\n
So,<\/p>\n
\\[
\ny’=\\frac{6\\cdot \\ln(3\\cdot x+1)}{3\\cdot x+1}.
\n\\]<\/p>\n
Example 4. Find the derivative of the function \\( y=(x^2+1)\\cdot \\ln^2(x) \\)<\/h3>\n
Again, we have a product, so we use the product rule. Let<\/p>\n
\\[
\nu=x^2+1,\\quad v=\\ln^2(x).
\n\\]<\/p>\n
Then<\/p>\n
\\[
\ny’=u’\\cdot v+u\\cdot v’.
\n\\]<\/p>\n
First, let us find the derivative of the first factor:<\/p>\n
\\[
\nu’=(x^2+1)’=2\\cdot x.
\n\\]<\/p>\n
Now let us find the derivative of the second factor. Here we again see the familiar part \\( \\ln^2(x) \\), so we use the formula we already learned:<\/p>\n
\\[
\nv’=\\bigl(\\ln^2(x)\\bigr)’=\\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
Substitute everything into the product rule:<\/p>\n
\\[
\ny’=2\\cdot x\\cdot \\ln^2(x)+(x^2+1)\\cdot \\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
So,<\/p>\n
\\[
\ny’=2\\cdot x\\cdot \\ln^2(x)+\\frac{2\\cdot (x^2+1)\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
This is the correct derivative of the given function.<\/p>\n
Example 5. Find the derivative of the function \\( y=\\bigl(\\ln^2(x)+1\\bigr)^3 \\)<\/h3>\n
Here we have a composite function in which the outer part is a cube, and the inner part is the expression \\( \\ln^2(x)+1 \\). In examples like this, it is especially important to move step by step from the outer part to the inner part.<\/p>\n
Let us denote<\/p>\n
\\[
\nu=\\ln^2(x)+1.
\n\\]<\/p>\n
Then<\/p>\n
\\[
\ny=u^3.
\n\\]<\/p>\n
The derivative of the outer part is<\/p>\n
\\[
\ny’=3\\cdot u^2\\cdot \\frac{du}{dx}.
\n\\]<\/p>\n
Now let us find the derivative of the inner expression:<\/p>\n
\\[
\n\\frac{du}{dx}=\\bigl(\\ln^2(x)+1\\bigr)’=\\bigl(\\ln^2(x)\\bigr)’+(1)’.
\n\\]<\/p>\n
The derivative of a constant is zero, and for \\( \\bigl(\\ln^2(x)\\bigr)’ \\) we use the formula we already know:<\/p>\n
\\[
\n\\frac{du}{dx}=\\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
After that, we return to the main expression:<\/p>\n
\\[
\ny’=3\\cdot \\bigl(\\ln^2(x)+1\\bigr)^2\\cdot \\frac{2\\cdot \\ln(x)}{x}.
\n\\]<\/p>\n
Multiply the numerical coefficients. Thus,<\/p>\n
\\[
\ny’=\\frac{6\\cdot \\ln(x)\\cdot \\bigl(\\ln^2(x)+1\\bigr)^2}{x}.
\n\\]<\/p>\n
What to Read Next: Recommended Topics to Continue With<\/h2>\n
Now that the topic of the derivative of the natural logarithm squared has been covered from the main formula to practical examples, it is quite natural to move on to related materials. Why is this useful? Because such topics complement one another well and help you see more clearly how different differentiation rules work in typical expressions.<\/p>\n
- \n
- Derivative of the Natural Logarithm: Formula, Proof, Examples<\/a> \u2014 This article focuses on the basic derivative of the logarithm, its derivation, and typical application problems.<\/li>\n
- Derivative of the Exponential Function: Formula, Proof, Examples<\/a> \u2014 Here, the derivative of the exponential function, its properties, and examples related to differentiation are discussed.<\/li>\n
- Derivative of the Square Root: Formula, Proof, Examples<\/a> \u2014 This topic explains how to find the derivative of the square root and how to apply it in different expressions.<\/li>\n<\/ol>\n
Derivative of the Natural Logarithm Squared: From Formula to Program Code<\/h2>\n
Now try to look at this topic not only as a mathematical rule, but also as the basis for a small programming project. Why not implement the flowchart shown below in your favorite programming language and check how your knowledge of the derivative of the natural logarithm squared works in real code? This kind of practice shows very well how a formula gradually turns into a sequence of conditions, calculations, and user messages. It also gives you a great chance to combine mathematical analysis with programming in a simple but meaningful task.<\/p>\n
![\"Flowchart](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/04\/derivative-of-the-natural-logarithm-squared2.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"The derivative of the natural logarithm squared is an important topic in mathematical analysis because it helps us better understand<\/p>\n","protected":false},"author":1,"featured_media":3228,"comment_status":"open","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[358],"tags":[363,499,501,496,500],"class_list":["post-3206","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-derivative-and-differential","tag-derivative-examples","tag-derivative-of-the-natural-logarithm-squared","tag-logarithmic-derivatives","tag-logarithmic-functions","tag-natural-logarithm-squared-derivative-proof"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=3206"}],"version-history":[{"count":20,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3206\/revisions"}],"predecessor-version":[{"id":3464,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/3206\/revisions\/3464"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/3228"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=3206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=3206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=3206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Derivative of the Exponential Function: Formula, Proof, Examples<\/a> \u2014 Here, the derivative of the exponential function, its properties, and examples related to differentiation are discussed.<\/li>\n