The derivative of arccot is a natural next step when you study inverse trigonometric functions and want not only to remember the result, but also to understand where it comes from. The function \( \operatorname{arccot}(x) \) does not behave like \( \operatorname{arctan}(x) \): it is decreasing, it has characteristic limiting values, and you can clearly see this behavior on a graph. Why does the derivative turn out to be negative? Why does the familiar expression \( 1+x^2 \) appear in the denominator again? And how does all of this follow from the definition of the derivative through a limit? Next, we will write down the formula, compare it with graphs, and then carefully derive the result from the limit definition of the derivative and reinforce everything with examples.
Derivative of Arccot: The Main Formula and What It Says About Decreasing Behavior
Let’s start with the key expression that the whole topic is built around. Let \( y=\operatorname{arccot}(x) \). Then its derivative is:
\[
\frac{d}{dx}\operatorname{arccot}(x)=-\frac{1}{1+x^2}.
\]
This formula works for all real \( x \). And right away it shows something important. First, the derivative is negative everywhere because \( 1+x^2>0 \) for any \( x \). So \( \operatorname{arccot}(x) \) decreases on the entire real line. Second, when \( |x| \) grows, the denominator \( 1+x^2 \) becomes larger, and the derivative approaches zero. That means the function changes more “slowly” far away from zero. Isn’t it interesting how a single formula immediately hints at the overall shape of the graph?
Below is an image of the graph of the function \( f(x)=\operatorname{arccot}(x) \) and its derivative \( f'(x)=-\frac{1}{1+x^2} \) in the same coordinate system.
From the figure, you can clearly see that \( \operatorname{arccot}(x) \) passes through the value \( \pi/2 \) at \( x=0 \), and then decreases: as \( x\to +\infty \) it tends to \( 0 \), and as \( x\to -\infty \) it tends to \( \pi \) (with the standard choice \( \operatorname{arccot}(x)\in(0,\pi)) \). The derivative, in turn, has its smallest value \( -1 \) at \( x=0 \) and then rises toward zero, but remains negative. So the sign of the derivative, the shape of the curve, and the limiting behavior all agree with each other.
Proof of the Formula: How the Derivative of Arccot Follows from the Definition
Now let’s move on to the derivation. If you want to feel truly confident using the derivative of arccot, it helps to see how the formula appears directly from the definition of the derivative—step by step, through clear transformations and standard limits.
Starting from the definition of the derivative
We begin with the basic definition. For \( y=\operatorname{arccot}(x) \), we have:
\[
\frac{d}{dx}\operatorname{arccot}(x)=\lim_{h\to 0}\frac{\operatorname{arccot}(x+h)-\operatorname{arccot}(x)}{h}.
\]
In the numerator we see the change in the function value, and in the denominator the change in the input, \( h \). The main task is to rewrite this fraction into a form where familiar trigonometric identities and well-known limits can be applied easily.
Introducing notation and switching to cotangent
To keep the writing compact, let’s set \( A=\operatorname{arccot}(x) \) and \(B=\operatorname{arccot}(x+h) \). Then the numerator becomes the difference \( B-A \).
Next, we use the meaning of an inverse function. If \( A=\operatorname{arccot}(x) \), then \( \cot(A)=x \). Similarly, if \( B=\operatorname{arccot}(x+h) \), then \( \cot(B)=x+h \). This means the increment of the argument can be written as a difference of cotangents: \( h=(x+h)-x=\cot(B)-\cot(A) \). So the expression inside the limit becomes:
\[
\frac{\operatorname{arccot}(x+h)-\operatorname{arccot}(x)}{h}
=\frac{B-A}{\cot(B)-\cot(A)}.
\]
This is the fraction we now need to analyze as \( h\to 0 \).
Turning the difference of cotangents into a single fraction
The next step is to simplify the denominator carefully. Expand the difference:
\[
\cot(B)-\cot(A)=\frac{\cos(B)}{\sin(B)}-\frac{\cos(A)}{\sin(A)}
=\frac{\cos(B) \cdot \sin(A)-\cos(A) \cdot \sin(B)}{\sin(A) \cdot \sin(B)}.
\]
In the numerator we recognize the expression \( \sin(A) \cdot \cos(B)-\cos(A) \cdot \sin(B) \), which is exactly the sine difference identity:
\[
\sin(A-B)=\sin(A) \cdot \cos(B)-\cos(A) \cdot \sin(B).
\]
Therefore,
\[
\cot(B)-\cot(A)=\frac{\sin(A-B)}{\sin(A) \cdot \sin(B)}.
\]
Substituting this into the earlier fraction gives:
\[
\frac{B-A}{\cot(B)-\cot(A)}
=\frac{B-A}{\dfrac{\sin(A-B)}{\sin(A) \cdot \sin(B)}}
=(B-A) \cdot \frac{\sin(A) \cdot \sin(B)}{\sin(A-B)}.
\]
The key point is that we rewrote the cotangent difference so that a sine of an angle difference appears in the denominator. This is what leads us to the standard limit involving \( \frac{\sin t}{t} \).
The key limit and an intermediate result
Now introduce \( t=B-A \). When \( h\to 0 \), the point \( x+h \) approaches \( x \), so \( \operatorname{arccot}(x+h)\to \operatorname{arccot}(x) \), meaning \( B\to A \). Hence \( t=B-A\to 0 \).
Also, notice that \( \sin(A-B)=\sin(-(B-A))=\sin(-t)=-\sin(t) \). So the expression becomes:
\[
(B-A) \cdot \frac{\sin(A) \cdot \sin(B)}{\sin(A-B)}
=t \cdot \frac{\sin(A) \cdot \sin(B)}{-\sin(t)}
=-\left(\sin(A) \cdot \sin(B)\right) \cdot \frac{t}{\sin(t)}.
\]
Therefore the derivative is:
\[
\frac{d}{dx}\operatorname{arccot}(x)
=\lim_{h\to 0}\left(-\left(\sin(A) \cdot \sin(B)\right) \cdot \frac{t}{\sin(t)}\right).
\]
Now we look at each factor in the limit. As \( h\to 0 \), we have \( B\to A \), so \( \sin(B)\to \sin(A) \), and the product \( \sin(A) \cdot \sin(B)\to \sin^2(A) \). At the same time \( t\to 0 \), and we can use the well-known limit \( \lim_{t\to 0}\frac{\sin(t)}{t}=1 \), which implies \( \lim_{t\to 0}\frac{t}{\sin(t)}=1 \). This gives the intermediate result:
\[
\frac{d}{dx}\operatorname{arccot}(x)=-\sin^2(A).
\]
Returning from the angle to the variable x
All that remains is to express \( \sin^2(A) \) in terms of \( x \). We know \( \cot(A)=x \). Use the identity \( 1+\cot^2(A)=\csc^2(A)=\frac{1}{\sin^2(A)} \). So \( \sin^2(A)=\frac{1}{1+\cot^2(A)} \). Substituting \( \cot(A)=x \), we get \( \sin^2(A)=\frac{1}{1+x^2} \).
Now substitute this into the intermediate result \( \frac{d}{dx}\operatorname{arccot}(x)=-\sin^2(A) \), and we obtain the final formula:
\[
\frac{d}{dx}\operatorname{arccot}(x)=-\frac{1}{1+x^2}.
\]
So we have derived the desired formula directly from the definition of the derivative, consistently using trigonometric identities and standard limits while keeping the logical chain clear at every step.
Practical Section: Derivative of Arccot in Examples
Theory gives you the big picture, but real confidence comes when you start differentiating specific functions on your own. In this section, we’ll go through \( 5 \) typical problems where the derivative of arccot appears together with the chain rule, the product rule, and the quotient rule. Before looking at each solution, try to find the derivative yourself—this is one of the best ways to make the result stick.
Example 1. Find the derivative of \( f(x)=\operatorname{arccot}(3 \cdot x) \)
Here we have a composite function. The outer function is \( g(u)=\operatorname{arccot}(u) \), and the inner function is \( u=3 \cdot x \). By the chain rule, we first differentiate the outer function (keeping the argument \( u \) unchanged):
\[
g'(u)=-\frac{1}{1+u^2}.
\]
Then we multiply by the derivative of the inner function: \( u’=3 \). So,
\[
f'(x)=g'(3 \cdot x) \cdot 3=-\frac{1}{1+(3 \cdot x)^2} \cdot 3=-\frac{3}{1+9 \cdot x^2}.
\]
So the derivative of \( \operatorname{arccot}(3 \cdot x) \) is \( f'(x)=-\frac{3}{1+9 \cdot x^2} \).
Example 2. Find the derivative of \( f(x)=x \cdot \operatorname{arccot}(x) \)
This time we have a product of two functions, so we use the product rule. Let \( u=x \) and \( v=\operatorname{arccot}(x) \). Then \( u’=1 \) and \( v’=-\frac{1}{1+x^2} \). Using \( (u \cdot v)’=u’ \cdot v+u \cdot v’ \), we get:
\[
f'(x)=1 \cdot \operatorname{arccot}(x)+x \cdot \left(-\frac{1}{1+x^2}\right)
=\operatorname{arccot}(x)-\frac{x}{1+x^2}.
\]
Final answer: \( f'(x)=\operatorname{arccot}(x)-\frac{x}{1+x^2} \).
Example 3. Find the derivative of \( f(x)=\big(\operatorname{arccot}(2 \cdot x)\big)^2 \)
Here we see a composition: the outer function is a square, and inside it we have \( \operatorname{arccot}(2 \cdot x) \). It’s convenient to work “from outside to inside”. First, differentiate the square:
\[
\frac{d}{dx}\big(\operatorname{arccot}(2 \cdot x)\big)^2
=2 \cdot \operatorname{arccot}(2 \cdot x) \cdot \frac{d}{dx}\big(\operatorname{arccot}(2 \cdot x)\big).
\]
Now find the derivative of \( \operatorname{arccot}(2 \cdot x) \). This is again the chain rule: the derivative of \( \operatorname{arccot}(u) \) is \( -\frac{1}{1+u^2} \), and for \( u=2 \cdot x \) we have \( u’=2 \). Therefore,
\[
\frac{d}{dx}\big(\operatorname{arccot}(2 \cdot x)\big)
=-\frac{1}{1+(2 \cdot x)^2} \cdot 2=-\frac{2}{1+4 \cdot x^2}.
\]
Now return to the original derivative:
\[
f'(x)=2 \cdot \operatorname{arccot}(2 \cdot x) \cdot \left(-\frac{2}{1+4 \cdot x^2}\right)
=-\frac{4 \cdot \operatorname{arccot}(2 \cdot x)}{1+4 \cdot x^2}.
\]
Final answer: \( f'(x)=-\frac{4 \cdot \operatorname{arccot}(2 \cdot x)}{1+4 \cdot x^2} \).
Example 4. Find the derivative of \( f(x)=\dfrac{\operatorname{arccot}(x)}{1+x^2} \)
Now we have a quotient, so we use the quotient rule. Let \( u=\operatorname{arccot}(x) \) 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 obtain:
\[
f'(x)=\frac{\left(-\frac{1}{1+x^2}\right) \cdot (1+x^2)-\operatorname{arccot}(x) \cdot 2 \cdot x}{(1+x^2)^2}.
\]
In the first term, \( (1+x^2) \) cancels out, so:
\[
f'(x)=\frac{-1-2 \cdot x \cdot \operatorname{arccot}(x)}{(1+x^2)^2}.
\]
This is already quite compact, so it’s convenient to leave it in this form.
Example 5. Find the derivative of \( f(x)=e^{2 \cdot x} \cdot \operatorname{arccot}(x) \)
Here we again have a product of two functions, but one of them is exponential. Let \( u=e^{2 \cdot x} \) and \( v=\operatorname{arccot}(x) \). For \( u \), use the chain rule: the derivative of \( e^{2 \cdot x} \) is \( 2 \cdot e^{2 \cdot x} \). For \( v \), we use the familiar derivative \( v’=-\frac{1}{1+x^2} \).
Now apply the product rule:
\[
f'(x)=2 \cdot e^{2 \cdot x} \cdot \operatorname{arccot}(x)+e^{2 \cdot x} \cdot \left(-\frac{1}{1+x^2}\right).
\]
It’s convenient to factor out \( e^{2 \cdot x} \) to make the answer look cleaner:
\[
f'(x)=e^{2 \cdot x} \cdot \left(2 \cdot \operatorname{arccot}(x)-\frac{1}{1+x^2}\right).
\]
That is the final answer.
What’s Next: Where to Go After the Topic “Derivative of Arccot”?
Do you want more than just memorizing the formula—do you want to feel that you truly control the topic? Then the next logical step is to move on to derivatives of related inverse trigonometric functions. This way, you’ll start recognizing common techniques in problems more quickly and you’ll feel more confident working with different types of examples.
- Derivative of Arcsin: Formula, Proof, Examples — In this article, we break down how to find the derivative of arcsine, where the formula comes from, and how to apply it in typical problems.
- Derivative of Arccos: Formula, Proof, Examples — We discuss the derivative of arccosine, its sign and behavior, and then solve several examples with detailed explanations.
- Derivative of Arctan: Formula, Proof, Examples — You’ll learn how the derivative of arctangent works, how it connects to the behavior of the graph, and how to solve practical exercises efficiently.
If you’re already practicing derivatives actively but sometimes feel unsure about your answer, it can be helpful to use an online derivative calculator to quickly check your computations.
Derivative of Arccot: From a Math Idea to Your Own Implementation
Now that you’ve worked through the derivative of arccotangent, it’s a perfect moment to turn this knowledge into a small—but genuinely practical—project. Take the flowchart shown below and implement it in your favorite language: Python, JavaScript, C#, Java, or even Pascal. Imagine you’re building a mini tool: it loops through values of \( x \) on a chosen interval, computes the derivative, and finds the point where the function decreases the fastest. This is great practice for loops, numerical accuracy, and clean result formatting.
And there’s an even more interesting part: you can check whether the output always matches what the mathematics suggests. Where exactly does the derivative reach its smallest value? And how does that depend on the parameter \( k \)? Tasks like this help you feel that calculus and programming work together instead of living in separate worlds.
