The derivative of arccos is a topic that connects trigonometry, inverse functions, and the basic ideas of calculus. It describes how the value of the function \( \arccos(x) \) changes when its input changes, and how this change can be captured using a precise analytical formula. Why does a negative sign always appear in this derivative? Where does the square root in the denominator come from? And how is all of this related to the graph of the function? In this article, we will first write down the main formula, then carefully derive it from the definition of the derivative, and finally look at some practical examples.
Main Formula: What Does the Derivative of Arccos Tell Us?
Let’s start with the main formula we want to understand and prove. For the function \( y = \arccos(x) \), its derivative has the form:
\[
\frac{d}{dx}\bigl(\arccos(x)\bigr) = -\frac{1}{\sqrt{1 – x^2}};
\]
This formula holds for all \( x \in (-1, 1) \). Although the function \( \arccos(x) \) itself is defined on the closed interval \( [-1, 1] \), the derivative does not exist at the endpoints. When \( x = -1 \) or \( x=1 \), the denominator tends to zero, and the value of the derivative becomes unbounded. So, the domain of the derivative is narrower than the domain of the original function.
A key conclusion from this formula is the sign of the derivative. It is always negative on its domain. This means that the function \( \arccos(x) \) is decreasing: as \( x \) increases, the value of the arccosine decreases. Let’s look at this from a geometric point of view. Imagine a single diagram where we plot both the graph of \( f(x) = \arccos(x) \) and the graph of its derivative \( f'(x) = -\frac{1}{\sqrt{1 – x^2}} \).
On this graph, you can clearly see that the curve \( \arccos(x) \) decreases monotonically on the whole interval \( [-1, 1] \). Near the middle, close to zero, it goes down more gently, while near the points \( x = -1 \) and \( x=1 \) the descent becomes much steeper. The derivative reflects this behavior very clearly: its values are always below zero, and their absolute value increases as \( x \) approaches the edges of the interval. In this way, the algebraic formula, the sign of the derivative, and the shape of the graph are in complete agreement.
Proof of the Formula: How Does the Derivative of Arccos Follow from the Definition?
Now let’s see how this formula is obtained using only the definition of the derivative. We start with the classical definition. For the function \( \arccos(x) \) we have:
\[
\frac{d}{dx}\bigl(\arccos(x)\bigr) = \lim_{h \to 0} \frac{\arccos(x+h) – \arccos(x)}{h};
\]
At first glance, this expression looks not very convenient: in the numerator we have the difference of two arccosine values, and it is not easy to work directly with such a fraction. So we introduce some auxiliary notation to simplify the expression and better see its structure.
Let \( A = \arccos(x) \), \( B = \arccos(x+h) \). Then, by the definition of arccosine, we have \( \cos(A) = x \) and \( \cos(B) = x + h \). So the difference in the numerator, that is \( \arccos(x+h) – \arccos(x) \), becomes simply \( B – A \).
Next, we want to get rid of\( h \) in the denominator. By our notation, \( h = (x + h) – x = \cos(B) – \cos(A) \). Therefore, the fraction can be written as:
\[
\frac{B – A}{h} = \frac{B – A}{\cos(B) – \cos(A)};
\]
So the problem reduces to finding the limit of a fraction whose denominator is a difference of cosines. At this point, it is natural to use a well-known trigonometric identity.
Recall the formula for the difference of cosines:
\[
\cos(B) – \cos(A) = -2 \cdot \sin\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right);
\]
Substitute this into the denominator:
\[
\frac{B – A}{\cos(B) – \cos(A)} = \frac{B – A}{-2 \cdot \sin\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right)};
\]
To make the structure clearer, rewrite the numerator as \( B – A = 2 \cdot \frac{B – A}{2} \). Then we have:
\[
\frac{B – A}{\cos(B) – \cos(A)} = \frac{2 \cdot \frac{B – A}{2}}{-2 \cdot \sin\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right)}
= -\frac{\frac{B – A}{2}}{\sin\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right)};
\]
Now it is convenient to separate this into two factors:
\[
\frac{B – A}{\cos(B) – \cos(A)} = -\left(\frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)}\right) \cdot \left(\frac{1}{\sin\left(\frac{A + B}{2}\right)}\right);
\]
In this form, it is much easier to study the limit. Next, we analyze the behavior of each factor separately as \( h \to 0 \).
First Factor — Moving to a Standard Limit
When \( h \to 0 \), we have \( B = \arccos(x + h) \to \arccos(x) = A \), that is, \( B \to A \). Accordingly, the difference \( B – A \to 0 \), and therefore \( \frac{B – A}{2} \to 0 \).
To make the notation more compact, introduce a temporary variable \( t = \frac{B – A}{2} \). Then the first factor can be written as
\[
\frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)} = \frac{t}{\sin (t)};
\]
Now we use the classical limit \( \lim_{t \to 0} \frac{\sin (t)}{t} = 1 \). From this it follows that \( \lim_{t \to 0} \frac{t}{\sin (t)} = 1 \). So the first factor approaches \( 1 \) in the limit.
Second Factor — Moving to \( sin(A) \)
Now consider the second factor \( \frac{1}{\sin\left(\frac{A + B}{2}\right)} \). When \( h \to 0 \), we again have \( B \to A \), so the average of the angles \( \frac{A + B}{2} \to \frac{A + A}{2} = A \).
Sine is a continuous function, so \( \sin\left(\frac{A + B}{2}\right) \to \sin(A) \), and therefore \( \frac{1}{\sin\left(\frac{A + B}{2}\right)} \to \frac{1}{\sin(A)} \).
So, in the limit, the second factor becomes \( \frac{1}{\sin(A)} \).
Combining the Results — An Intermediate Formula
Now combine both results. We have:
\[
\lim_{h \to 0} \frac{B – A}{\cos(B) – \cos(A)}
= \lim_{h \to 0} \left(-\left(\frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)}\right)\cdot\left(\frac{1}{\sin\left(\frac{A + B}{2}\right)}\right)\right)
= -1 \cdot \frac{1}{\sin(A)} = -\frac{1}{\sin(A)};
\]
We also remember that
\[
\frac{d}{dx}\bigl(\arccos(x)\bigr) = \lim_{h \to 0} \frac{B – A}{h}
= \lim_{h \to 0} \frac{B – A}{\cos(B) – \cos(A)};
\]
Therefore, we obtain the intermediate formula \( \frac{d}{dx}\bigl(\arccos(x)\bigr) = -\frac{1}{\sin(A)} \), where \( A = \arccos(x) \). All that remains is to express \( \sin(A) \) in terms of \( x \), so that the derivative is written completely using the variable \( x \).
Expressing \( sin(A) \) in Terms of \( x \) and the Final Form of the Derivative of Arccos
From our earlier notation we have \( \cos(A) = x \) and \( A = \arccos(x) \). The sine and cosine functions are connected by the fundamental trigonometric identity:
\[
\sin^2(A) + \cos^2(A) = 1;
\]
Substituting \( \cos(A) = x \), we get \( \sin^2(A) + x^2 = 1 \), so \( \sin^2(A) = 1 – x^2 \).
Now recall that \( A = \arccos(x) \in [0, \pi] \). On this interval, sine is nonnegative, that is \( \sin(A) \ge 0 \). Therefore, we take the positive square root: \( \sin(A) = \sqrt{1 – x^2} \).
Substitute this expression into the intermediate formula for the derivative:
\[
\frac{d}{dx}\bigl(\arccos(x)\bigr) = -\frac{1}{\sin(A)} = -\frac{1}{\sqrt{1 – x^2}};
\]
So we have rigorously derived the formula for the derivative of arccos, using only the definition of the derivative, trigonometric identities, and properties of limits. The step-by-step analysis shows how each transformation logically leads to the final result and explains why the derivative of \( \arccos(x) \) has exactly this form.
Practical Block: Derivative of Arccos in Examples
To use the formula where the derivative of arccos appears with confidence, it’s important not only to understand the theory but also to practice on specific functions. This way, you can see how the chain rule, product rule, and quotient rule work, and how composite expressions behave in real calculations. Before reading the solutions, try to find the derivatives on your own — this really helps to strengthen your understanding.
Example 1: Find the derivative of \( f(x) = \arccos(3 \cdot x) \)
Here we have a classic composite function. The outer part is \( g(u) = \arccos(u) \), and the inner part is \( u = 3 \cdot x \). By the chain rule, we first differentiate the outer function, keeping the argument \( u \) unchanged: \( g'(u) = – \frac{1}{\sqrt{1 – u^2}} \). Then we multiply by the derivative of the inner function, which is \( u’ = 3 \). We get:
\[
f'(x) = \left(-\frac{1}{\sqrt{1 – (3x)^2}}\right) \cdot 3 = -\frac{3}{\sqrt{1 – 9x^2}};
\]
So, the final result is: \( f'(x) = -\frac{3}{\sqrt{1 – 9x^2}} \).
Example 2: Find the derivative of \( f(x) = x \cdot \arccos(x) \)
In this example we have a product of two functions, so we use the product rule. Let \( u = x \) and \( v = \arccos(x) \). Then \( u’ = 1 \) and \( v’ = – \frac{1}{\sqrt{1 – x^2}} \). Using the formula \( (u \cdot v)’ = u’ \cdot v + u \cdot v’ \), we have:
\[
f'(x) = 1 \cdot \arccos(x) + x \cdot \left(-\frac{1}{\sqrt{1 – x^2}}\right)
= \arccos(x) – \frac{x}{\sqrt{1 – x^2}};
\]
So, the final result is: \( f'(x) = \arccos(x) – \frac{x}{\sqrt{1 – x^2}} \).
Example 3: Find the derivative of \( f(x) = \bigl(\arccos(2 \cdot x)\bigr)^2 \)
Here we have a composite expression: first a square, then the arccos function, and inside it a linear function \( 2 \cdot x \). We move from the outside inward.
First, differentiate the square:
\[
\frac{d}{dx}(\arccos(2 \cdot x))^2 = 2 \cdot \arccos(2 \cdot x) \cdot \frac{d}{dx}(\arccos(2 \cdot x));
\]
Next, we apply the chain rule to \( \arccos(2 \cdot x) \). The outer function is \( \arccos(u) \) with derivative \( – \frac{1}{\sqrt{1 – u^2}} \), and the inner function is \( u = 2 \cdot x \) with derivative \( u’ = 2 \):
\[
\frac{d}{dx}(\arccos(2 \cdot x)) = -\frac{1}{\sqrt{1 – (2 \cdot x)^2}} \cdot 2 = -\frac{2}{\sqrt{1 – 4 \cdot x^2}};
\]
Now we combine both steps:
\[
f'(x) = 2 \cdot \arccos(2 \cdot x) \cdot \left(-\frac{2}{\sqrt{1 – 4 \cdot x^2}}\right)
= -\frac{4 \cdot \arccos(2 \cdot x)}{\sqrt{1 – 4 \cdot x^2}};
\]
So, the final result is: \( f'(x) = -\frac{4 \cdot \arccos(2 \cdot x)}{\sqrt{1 – 4 \cdot x^2}} \).
Example 4: Find the derivative of \( f(x) = \dfrac{\arccos(x)}{1 + x^2} \)
Now we have a quotient, so we use the quotient rule. Let \( u = \arccos(x) \), \( v = 1 + x^2 \). Then \( u’ = – \frac{1}{\sqrt{1 – x^2}} \) and \( v’ = 2 \cdot x \). Using the formula \( \left(\frac{u}{v}\right)’ = \frac{u’ \cdot v – u \cdot v’}{v^2} \), we obtain:
\[
f'(x) = \frac{\left(-\frac{1}{\sqrt{1 – x^2}}\right) \cdot (1 + x^2) – \arccos(x) \cdot 2 \cdot x}{(1 + x^2)^2};
\]
If desired, we can just rearrange the numerator without changing the essence:
\[
f'(x) = \frac{-\frac{1 + x^2}{\sqrt{1 – x^2}} – 2 \cdot x \cdot \arccos(x)}{(1 + x^2)^2};
\]
This is the complete answer.
Example 5: Find the derivative of \( f(x) = e^{2 \cdot x} \cdot \arccos(x) \)
Again we have a product, so we apply the product rule. Let \( u = e^{2 \cdot x} \) and \( v = \arccos(x) \). For \( u \), we use the chain rule: \( u’ = 2 \cdot e^{2 \cdot x} \). For \( v \), we have \( v’ = – \frac{1}{\sqrt{1 – x^2}} \). Combine them using \( (u \cdot v)’ = u’ \cdot v + u \cdot v’ \):
\[
f'(x) = 2 \cdot e^{2 \cdot x} \cdot \arccos(x) + e^{2 \cdot x} \cdot \left(-\frac{1}{\sqrt{1 – x^2}}\right);
\]
It is convenient to factor out the common factor \( e^{2 \cdot x} \):
\[
f'(x) = e^{2 \cdot x} \cdot \left(2 \cdot \arccos(x) – \frac{1}{\sqrt{1 – x^2}}\right);
\]
So, the final result is: \( f'(x) = e^{2 \cdot x} \cdot \left(2 \cdot \arccos(x) – \frac{1}{\sqrt{1 – x^2}}\right) \).
What Comes After the Derivative of Arccos? Recommended Topics to Continue
If you already understand the derivative of arccos, it’s a great moment to widen the picture and look at other inverse trigonometric functions. Each of them has its own features, but together they form a complete system that is very important in calculus and many applied problems. Below are topics that are convenient to study next.
- Derivative of Arcsine: Formula, Proof, Examples — This article explains how the derivative of arcsin is obtained and shows how to use it through several common examples.
- Derivative of Arctangent: Formula, Proof, Examples — This material reveals the key properties of the derivative of arctangent and shows how to apply it in calculations and analysis problems.
- Derivative of Arccotangent: Formula, Proof, Examples — The article presents the derivation of the derivative of arccotangent and demonstrates how it helps to solve problems of different difficulty levels.
If you are already practicing derivatives but sometimes hesitate about your answer, you can use an online derivative calculator to quickly check your computations.
Derivative of Arccos in Code: From Flowchart to Working Program
If you enjoy programming and like seeing how mathematics comes to life in code, pay attention to the flowchart of the algorithm for finding the equation of the tangent to the graph of the arccosine function at a point chosen by the user. Using this flowchart, you can create your own program — for example, in Pascal or any other language you find convenient — that step by step reads the input data, performs the calculations, and outputs the equation of the tangent in a clear form.
This task not only reinforces your understanding of how the derivative of arccos works, but also trains your ability to turn a mathematical algorithm into a real software product. Try implementing the flowchart on your own and see how logic from paper turns into a working program.
