The derivative of arctan is another important step when you study inverse trigonometric functions and calculus in general. It shows how the value of the function \( \arctan (x) \) changes when we slightly change its input, and how this change can be described by an exact formula. Why does a simple expression like “one plus \( x \) squared” suddenly appear in the denominator? How is this connected to the properties of the usual tangent? And what does the graph of the derivative tell us about how the function behaves? In this article, we will first write down the main formula and see how it looks on graphs, then derive it step by step from the definition of the derivative, and after that move on to practical examples.
Derivative of Arctan: The Main Formula and What It Says About Growth
Let’s start with the key expression that the whole topic is built around. Let \( y = arctan (x) \). Then its derivative is:
\[ \frac{d}{dx}\bigl(\arctan(x)\bigr) = \frac{1}{1 + x^2}; \]
This formula is true for all real \( x \). Unlike \( \arcsin(x) \) or \( \arccos(x) \), the function \( \arctan (x) \) is defined on the entire number line, and its derivative also exists for every \( x \in \mathbb{R} \). The denominator \( 1 + x^2 \) is never equal to zero, so there are no “problem” points here.
Right away, the formula shows a few important things. First, the derivative is always positive, because \( 1 + x^2 > 0 \) for any \( x \). So \( arctan (x) \) is increasing everywhere. Second, as \( |x| \) becomes larger, the denominator \( 1 + x^2 \) grows, and the derivative values become smaller. This means \( arctan (x) \) grows faster near zero and more slowly as \( x \) moves away from the origin. In this way, we can clearly see the connection between the algebraic formula and the behavior of the graph.
To picture it better, it helps to look at two graphs in the same coordinate system: \( f(x) = \arctan(x) \) and \( f'(x) = \frac{1}{1 + x^2} \).
From the picture, you can see that the curve \( arctan (x) \) has an S-shape: it slowly approaches the horizontal asymptote \( y = \frac{\pi}{2} \) as \( x \to +\infty \), and \( y = -\frac{\pi}{2} \) as \( x \to -\infty \). Near \( x = 0 \), the graph is steeper, meaning the function values change faster. The derivative graph shows this clearly: \( f'(x) = \frac{1}{1 + x^2} \) reaches its maximum value \( 1 \) at \( x = 0 \), and then smoothly decreases toward zero, while always staying positive. So the algebraic formula, the sign of the derivative, and the shape of the graphs all agree and give one clear picture of how the derivative of arctangent works.
Proving the Formula: How the Derivative of Arctan Follows from the Definition
Now let’s look at how the familiar formula for \( arctan (x) \) actually appears from the definition of the derivative. To truly understand the topic, it helps not only to remember the final result, but also to see the logical path from the basic definition all the way to the finished formula.
Starting from the definition of the derivative
We begin with the classic definition. For the function \( y = \arctan (x) \) we write:
\[ \frac{d}{dx}\bigl(\arctan(x)\bigr) = \lim_{h \to 0} \frac{\arctan(x+h) – \arctan(x)}{h}; \]
In this expression, the numerator contains the difference of two arctangent values, and that is exactly the part we need to handle carefully. To make the writing shorter and easier to work with, let’s introduce \( A = \arctan(x) \) and \( B = \arctan(x + h) \). Then, by the meaning of the inverse function, we have \( \tan(A) = x \) and \( \tan(B) = x + h \).
The difference in the numerator, \( \arctan(x + h) – \arctan(x) \), becomes \( B – A \). And the increment \( h \) in the denominator can be rewritten as \( h = (x + h) – x = \tan(B) – \tan(A) \). So the expression inside the limit turns into:
\[ \frac{\arctan(x + h) – \arctan(x)}{h} = \frac{B – A}{\tan(B) – \tan(A)}; \]
This is the fraction we must investigate as \( h \to 0 \). When \( h \) goes to zero, the point \( x+h \) approaches \( x \), so \( B \to A \), and the difference \( B-A \) becomes very small. That is exactly what the next step uses.
Moving to a tangent difference and changing variables
At this stage we use a known trigonometric identity for the difference of tangents. For any angles \( A \) and \( B \), we have:
\[ \tan(B) – \tan(A) = \frac{\sin(B – A)}{\cos(A) \cdot \cos(B)}; \]
This identity can be derived from \( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \), but the key point here is simple: it allows us to replace a tangent difference with an expression involving \( \sin(B – A) \).
Substituting this into the denominator gives:
\[ \frac{B – A}{\tan(B) – \tan(A)} = \frac{B – A}{\frac{\sin(B – A)}{\cos(A) \cdot \cos(B)}} = (B – A) \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(B – A)}; \]
To make the limit clearer, introduce a temporary variable \( t = B – A \). As \( h \to 0 \), we know \( B \to A \), so \( t = B – A \to 0 \). Then the expression becomes:
\[ (B – A) \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(B – A)} = t \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(t)}; \]
So we can rewrite the derivative as:
\[ \frac{d}{dx}\bigl(\arctan(x)\bigr) = \lim_{h \to 0} \frac{\arctan(x + h) – \arctan(x)}{h} = \lim_{h \to 0} t \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(t)}; \]
Evaluating the limit
When \( h \to 0 \), we have \( B \to A \), which means \( \cos(B) \to \cos(A) \). So the product \( \cos(A) \cdot \cos(B) \) approaches \( \cos^2(A) \).
At the same time, \( t \to 0 \), and here a basic calculus limit becomes useful \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \), so \( \lim_{t \to 0} \frac{t}{\sin(t)} = 1 \). Now split the expression into two parts:
\[ t \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(t)} = \left( \cos(A) \cdot \cos(B) \right) \cdot \frac{t}{\sin(t)}; \]
We can take the limit of each part. As \( h \to 0 \), \( B \to A \), тому \( \cos(A) \cdot \cos(B) \to \cos^2(A) \), and \( \frac{t}{\sin(t)} \to 1 \). Therefore,
\[ \lim_{h \to 0} t \cdot \frac{\cos(A) \cdot \cos(B)}{\sin(t)} = \cos^2(A) \cdot 1 = \cos^2(A); \]
So we get the intermediate result:
\[ \frac{d}{dx}\bigl(\arctan(x)\bigr) = \cos^2(A); \]
where \( A = \arctan(x) \). At this point the derivative is written in terms of the angle \( A \), and the final step is to rewrite it in terms of \( x \).
Switching from the angle back to x
To finish the proof, we need to connect \( \cos^2(A) \) with \( x \). We already know that \( \tan(A) = x \). From trigonometry, there is a standard identity that links tangent and cosine:
\[ 1 + \tan^2(A) = \frac{1}{\cos^2(A)}; \]
This can be rewritten as \( \cos^2(A) = \frac{1}{1 + \tan^2(A)} \). Now substitute \( \tan(A) = x \): \( \cos^2(A) = \frac{1}{1 + x^2} \). Return to the expression for the derivative and replace \( \cos^2(A) \):
\[ \frac{d}{dx}\bigl(\arctan(x)\bigr) = \frac{1}{1 + x^2}; \]
That is the exact formula we started with when discussing the derivative of arctangent. We reached it using only the definition of the derivative, trigonometric identities, and basic limit properties. Step by step, each transformation logically leads to the final result and explains why the derivative of \( \arctan(x) \) has this particular form.
Practical Section: The Derivative of Arctan in Real Examples
Theory gives you the big picture, but real understanding usually comes when you start solving specific problems. In this part, we’ll see how the derivative of arctangent works in practice and how the chain rule, the product rule, and the quotient rule show up in action. Before you read each solution, try to find the derivative on your own first — that habit really strengthens your skills.
Example 1: Find the derivative of \( f(x) = \arctan(3 \cdot x) \)
This is a typical composite function. The outer function is \( g(u) = \arctan(u) \), and the inner function is \( u = 3 \cdot x \). By the chain rule, we first differentiate the outer function while keeping its input \( u \) unchanged: \( g'(u) = \frac{1}{1 + u^2} \). Then we multiply by the derivative of the inner function, \( u' = 3 \).
So we get:
\[ f'(x) = \left( \frac{1}{1 + (3 \cdot x)^2} \right) \cdot 3 = \frac{3}{1 + 9 \cdot x^2}; \]
In summary, the derivative of \( \arctan(3 \cdot x) \) is \( f'(x) = \frac{3}{1 + 9 \cdot x^2} \).
Example 2: Find the derivative of \( f(x) = x \cdot \arctan(x) \)
Here we have a product of two functions, so we use the product rule. Let \( u = x \) and \( v = \arctan(x) \). Then \( u' = 1 \) and \( v' = \frac{1}{1 + x^2} \). Using \( (u \cdot v)' = u' \cdot v + u \cdot v' \), we obtain:
\[ f'(x) = 1 \cdot \arctan(x) + x \cdot \left( \frac{1}{1 + x^2} \right) = \arctan(x) + \frac{x}{1 + x^2}; \]
So the final result is \( f'(x) = \arctan(x) + \frac{x}{1 + x^2} \).
Example 3: Find the derivative of \( f(x) = \bigl(\arctan(2 \cdot x)\bigr)^2 \)
This is a layered composition: first a square, then an arctangent, and inside that a linear function \( 2 \cdot x \). It’s convenient to move “from the outside inward”. First differentiate the square:
\[ \frac{d}{dx} \left( \arctan(2 \cdot x) \right)^2 = 2 \cdot \arctan(2 \cdot x) \cdot \frac{d}{dx} \arctan(2 \cdot x); \]
Now apply the chain rule to \( \arctan(2 \cdot x) \). The outer function is \( \arctan(u) \) with derivative \( \frac{1}{1 + u^2} \), and the inner function is \( u = 2 \cdot x \) with \( u' = 2 \). So:
\[ \frac{d}{dx} \arctan(2 \cdot x) = \frac{1}{1 + (2 \cdot x)^2} \cdot 2 = \frac{2}{1 + 4 \cdot x^2}; \]
Substitute this back into the derivative of the original function:
\[ f'(x) = 2 \cdot \arctan(2 \cdot x) \cdot \frac{2}{1 + 4 \cdot x^2} = \frac{4 \cdot \arctan(2 \cdot x)}{1 + 4 \cdot x^2}; \]
So the final answer is \( f'(x) = \frac{4 \cdot \arctan(2 \cdot x)}{1 + 4 \cdot x^2} \).
Example 4: Find the derivative of \( f(x) = \dfrac{\arctan(x)}{1 + x^2} \)
Now we have a quotient, so we use the quotient rule. Let \( u = \arctan(x) \) and \( v = 1 + x^2 \). Then \( u' = \frac{1}{1 + x^2} \) and \( v' = 2 \cdot x \). Using \( \left( \frac{u}{v} \right)' = \frac{u' \cdot v – u \cdot v'}{v^2} \), we get:
\[ f'(x) = \frac{\left( \frac{1}{1 + x^2} \right) \cdot (1 + x^2) – \arctan(x) \cdot 2 \cdot x}{(1 + x^2)^2} = \frac{1 – 2 \cdot x \cdot \arctan(x)}{(1 + x^2)^2}; \]
This result is already compact, so it’s perfectly fine to leave it in this form.
Example 5: Find the derivative of \( f(x) = e^{2 \cdot x} \cdot \arctan(x) \)
Here we again have a product, but one factor is exponential. Let \( u = e^{2 \cdot x} \) and \( v = \arctan(x) \). For \( u \), apply the chain rule: \( u' = 2 \cdot e^{2 \cdot x} \). For \( v \), we use the familiar derivative \( v' = \frac{1}{1 + x^2} \).
Now apply the product rule \( (u \cdot v)' = u' \cdot v + u \cdot v' \):
\[ f'(x) = 2 \cdot e^{2 \cdot x} \cdot \arctan(x) + e^{2 \cdot x} \cdot \left( \frac{1}{1 + x^2} \right); \]
It’s convenient to factor out \( e^{2 \cdot x} \) to make the answer cleaner:
\[ f'(x) = e^{2 \cdot x} \cdot \left( 2 \cdot \arctan(x) + \frac{1}{1 + x^2} \right); \]
That is the final answer.
More Practice Ahead: What to Study After the Derivative of Arctan
Want to lock in the result and see the bigger picture? Then it’s a great idea to continue with other inverse trigonometric functions. These topics naturally fit right after the derivative of arctangent and help you build a more complete, confident understanding.
- Derivative of Arcsine: Formula, Proof, Examples — This article shows how to derive the derivative formula for arcsin and how to use it in typical calculus problems.
- Derivative of Arccosine: Formula, Proof, Examples — This material walks through the derivative of arccos from the main formula to the proof and clear, step-by-step examples.
- Derivative of Arccotangent: Formula, Proof, Examples — This article explains the formula and proof for the derivative of arccot and includes several examples to reinforce the theory.
If you are already practicing derivatives actively but sometimes doubt your final result, it can be helpful to use an online derivative calculator to quickly check your work.
Derivative of Arctan in Code: The Next Step for Programmers
If you enjoy programming and want to turn theory into real code, try implementing the algorithm from the flowchart that analyzes how the function \( \arctan (x) \) behaves on a user-selected interval. Since you already have a ready-made flowchart, you’ve covered half the journey: now you just need to translate its elements step by step into variables, condition checks, and loops in your favorite programming language. This kind of task not only helps you feel how the derivative of arctangent works in theory, but also shows how calculus is directly used to build programs that truly explore function behavior. If you’re ready for a challenge, take the flowchart, open your code editor, and try implementing the algorithm on your own.
