The derivative of arcsin is an important topic that brings together ideas from trigonometry, inverse functions, and basic mathematical analysis. By studying the derivative of this function, you can see how a small change in its argument affects the value of the inverse trigonometric function. This helps you learn how to apply mathematical concepts in practice when solving different types of problems.
In this article, we will carefully look at the main formula for the derivative and then derive it step by step from the definition of the derivative. This will help you understand why a square root appears in the denominator and what it means for the behavior of the function.
Main Formula: How the Derivative of Arcsin Behaves
Let’s start with the formula that we will later prove. For the function \( \arcsin(x) \), the derivative has the form:
\[
\frac{d}{dx}\bigl(\arcsin(x)\bigr) = \frac{1}{\sqrt{1 – x^2}};
\]
This formula is valid for all \( x \) in the interval \( (-1, 1) \). At the endpoints \( x = -1 \) and \( x = 1 \), the values of the function \( \arcsin(x) \) still exist, but the derivative “shoots off” to infinity, so it is not defined at those points. From this we get an important conclusion: the domain of the derivative is narrower than the domain of the original function.
Now let’s see what the graph tells us. The curve \( \arcsin(x) \) increases on the entire interval \( [-1, 1] \), but it does not increase in the same way everywhere: near zero it grows more gently, and closer to \( -1 \) and \( 1 \) the growth becomes much steeper. The derivative \( f'(x) = \frac{1}{\sqrt{1 – x^2}} \) shows this very clearly: it is positive on its whole domain, but it rises sharply as \( x \) approaches \( -1 \) or \( 1 \).
In this way, the algebraic formula and the geometric picture match perfectly: the fact that \( \arcsin(x) \) is increasing corresponds to the positivity of the derivative, and the steep slope of the graph near the endpoints corresponds to the large values of \( f'(x) \).
Proof via the Definition of the Derivative: Step by Step
Now let’s see how the formula for the derivative of \( \arcsin(x) \) is actually obtained. We start from the definition of the derivative. For the function \( y = \arcsin(x) \), it looks like this:
\[
\frac{d}{dx}\bigl(\arcsin(x)\bigr) = \lim_{h \to 0} \frac{\arcsin(x + h) – \arcsin(x)}{h};
\]
At first glance, this expression is quite inconvenient: in the numerator we have the difference of two arcsin values, and it is hard to work with that directly. So we take an important step — we introduce auxiliary notation to break everything into simpler pieces.
Let \( A = \arcsin(x) \), \( B = \arcsin(x + h) \). Then, from the definition of arcsine, we get two equalities: \( \sin(A) = x \), \( \sin(B) = x + h \). So the difference in the numerator of the original fraction is simply \( B – A \). Therefore, we can rewrite the definition of the derivative as:
\[
\frac{d}{dx}\bigl(\arcsin(x)\bigr) = \lim_{h \to 0} \frac{B – A}{h};
\]
Now express \( h \) through sines: \( h = (x + h) – x = \sin(B) – \sin(A) \). Thus,
\[
\frac{B – A}{h} = \frac{B – A}{\sin(B) – \sin(A)};
\]
So the problem reduces to computing the limit of a fraction where the denominator is the difference of two sine values.
We recall the trigonometric identity for the difference of sines:
\[
\sin(B) – \sin(A) = 2 \cdot \cos\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right);
\]
Substitute this formula into the denominator:
\[
\frac{B – A}{\sin(B) – \sin(A)} = \frac{B – A}{2 \cdot \cos\left(\frac{A + B}{2}\right) \cdot \sin\left(\frac{B – A}{2}\right)};
\]
To better see the structure of the expression, split the numerator in half \( B – A = 2 \cdot \frac{B – A}{2} \), and rewrite the fraction as
\[
\frac{B – A}{\sin(B) – \sin(A)}
= \frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)} \cdot \frac{1}{\cos\left(\frac{A + B}{2}\right)};
\]
In this form, it is much more convenient to study the limit.
Let’s look at the behavior of each factor separately as \( h \to 0 \). First, when \( h \to 0 \), we have \( B \to A \), because \( B = \arcsin(x + h) \) and \( A = \arcsin(x) \). Therefore, the difference \( B – A \) tends to zero, and so \( \frac{B – A}{2} \to 0 \).
Introduce a temporary variable \( t = \frac{B – A}{2} \). Then the first factor becomes:
\[
\frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)} = \frac{t}{\sin(t)};
\]
We know that \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \), so \( \lim_{t \to 0} \frac{t}{\sin(t)} = 1 \). Therefore, the first factor tends to \( 1 \) in the limit.
Next, consider the second factor \( \dfrac{1}{\cos\left(\frac{A + B}{2}\right)} \). When \( h \to 0 \), we have \( B \to A \), so the average value \( \frac{A + B}{2} \to \frac{A + A}{2} = A \). By the continuity of cosine, we immediately get \( \cos\left(\frac{A + B}{2}\right) \to \cos(A) \), and therefore \( \frac{1}{\cos\left(\frac{A + B}{2}\right)} \to \frac{1}{\cos(A)} \).
Now we can combine both results. In the limit we have:
\[
\lim_{h \to 0} \left(\frac{B – A}{\sin(B) – \sin(A)}\right)
= \lim_{h \to 0} \left(\frac{\frac{B – A}{2}}{\sin\left(\frac{B – A}{2}\right)}\right)
\cdot \lim_{h \to 0} \left(\frac{1}{\cos\left(\frac{A + B}{2}\right)}\right)
= 1 \cdot \frac{1}{\cos(A)} = \frac{1}{\cos(A)};
\]
So,
\[
\frac{d}{dx}\bigl(\arcsin(x)\bigr)
= \lim_{h \to 0} \frac{B – A}{h}
= \lim_{h \to 0} \frac{B – A}{\sin(B) – \sin(A)}
= \frac{1}{\cos(A)};
\]
It remains to express \( \cos(A) \) in terms of \( x \), so that the formula for the derivative of arcsin is written only using the variable \( x \). From our earlier notation we have \( \sin(A) = x \). For any angle \( A \), the fundamental trigonometric identity holds:
\[
\sin^2(A) + \cos^2(A) = 1;
\]
Substitute \( \sin(A) = x \) and get \( x^2 + \cos^2(A) = 1 \), so \( \cos^2(A) = 1 – x^2 \). Since \( A = \arcsin(x) \) belongs to the interval \( [\frac{-\pi}{2}, \frac{\pi}{2}] \), and on this interval cosine is nonnegative, we take the positive root \( \cos(A) = \sqrt{1 – x^2} \).
Now return to the expression for the derivative:
\[
\frac{d}{dx}\bigl(\arcsin(x)\bigr)
= \frac{1}{\cos(A)}
= \frac{1}{\sqrt{1 – x^2}};
\]
Thus, we have obtained the formula for the derivative of arcsin strictly from the definition of the derivative, using trigonometric identities, limits, and a change of variables. This step-by-step breakdown clearly shows how these mathematical transformations lead to this important formula.
Practice Block: Examples on the Topic of the Derivative of Arcsin
To confidently use the formula for the derivative of arcsin, it is very helpful to go through a few detailed examples. This way you will see how the chain rule works, how composite functions behave, and what typical algebraic simplifications look like. Before reading each solution, try to do the calculations on your own — this is a great way to strengthen your understanding.
Example 1: Find the derivative of \( f(x) = \arcsin(3 \cdot x) \)
Here we have a classic composite function: the outer part is \( g(u) = \arcsin(u) \), and the inner part is \( u = 3 \cdot x \). By the chain rule, we first differentiate the outer function \( g'(u) = \frac{1}{\sqrt{1 – u^2}} \), keeping \( u \) unchanged, and then multiply by the derivative of the inner part \( u’ = 3 \). We obtain:
\[
f'(x) = \frac{1}{\sqrt{1 – (3 \cdot x)^2}} \cdot 3
= \frac{3}{\sqrt{1 – 9 \cdot x^2}};
\]
So, the final result is \( f'(x) = \frac{3}{\sqrt{1 – 9 \cdot x^2}} \).
Example 2: Find the derivative of \( f(x) = x \arcsin(x) \)
Here we have a product of two functions, so we use the product rule. Let \( u = x \) and \( v = \arcsin(x) \). Then \( u’ = 1 \), \( v’ = \frac{1}{\sqrt{1 – x^2}} \). Substitute into the formula \( (u \cdot v)’ = u’ \cdot v + u \cdot v’ \):
\[
f'(x) = 1 \cdot \arcsin(x) + x \cdot \frac{1}{\sqrt{1 – x^2}}
= \arcsin(x) + \frac{x}{\sqrt{1 – x^2}};
\]
So, \( f'(x) = \arcsin(x) + \frac{x}{\sqrt{1 – x^2}} \).
Example 3: Find the derivative of \( f(x) = \bigl(\arcsin(2 \cdot x)\bigr)^2 \)
Here we have a composition “square → arcsin → linear function”. First, we differentiate the outer square:
\[
\frac{d}{dx}\bigl(\arcsin(2 \cdot x)\bigr)^2
= 2 \cdot \arcsin(2 \cdot x) \cdot \frac{d}{dx}\bigl(\arcsin(2 \cdot x)\bigr);
\]
Next, we use the chain rule for \( \arcsin(2 \cdot x) \):
\[
\frac{d}{dx}\bigl(\arcsin(2 \cdot x)\bigr)
= \frac{1}{\sqrt{1 – (2 \cdot x)^2}} \cdot 2
= \frac{2}{\sqrt{1 – 4 \cdot x^2}};
\]
Combining the steps, we get:
\[
f'(x) = 2 \cdot \arcsin(2 \cdot x) \cdot \frac{2}{\sqrt{1 – 4 \cdot x^2}}
= \frac{4 \cdot \arcsin(2 \cdot x)}{\sqrt{1 – 4 \cdot x^2}};
\]
So the final result is \( f'(x) = \frac{4 \cdot \arcsin(2 \cdot x)}{\sqrt{1 – 4 \cdot x^2}} \).
Example 4: Find the derivative of \( f(x) = \dfrac{\arcsin(x)}{1 + x^2} \)
We have a quotient, so we apply the quotient rule. Let \( u = \arcsin(x) \), \( v = 1 + x^2 \). Then \( u’ = \frac{1}{\sqrt{1 – x^2}} \), \( v’ = 2 \cdot x \). By the formula \( \left(\frac{u}{v}\right)’ = \frac{u’ \cdot v – u \cdot v’}{v^2} \) we obtain:
\[
f'(x) = \frac{\frac{1}{\sqrt{1 – x^2}} \cdot (1 + x^2) – \arcsin(x) \cdot 2 \cdot x}{(1 + x^2)^2};
\]
If needed, we can simply reorder the numerator without changing the essence of the calculation:
\[
f'(x) = \frac{(1 + x^2) \cdot \frac{1}{\sqrt{1 – x^2}} – 2 \cdot x \cdot \arcsin(x)}{(1 + x^2)^2};
\]
This is the complete answer.
Example 5: Find the derivative of \( f(x) = e^{2 \cdot x} \cdot \arcsin(x) \)
Again we have a product, so the product rule applies. Let \( u = e^{2 \cdot x} \) and \( v = \arcsin(x) \). 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 the results:
\[
f'(x) = 2 \cdot e^{2 \cdot x} \cdot \arcsin(x) + e^{2 \cdot x} \cdot \frac{1}{\sqrt{1 – x^2}};
\]
It is convenient to factor out the common factor \( e^{2 \cdot x} \):
\[
f'(x) = e^{2 \cdot x} \cdot \left(2 \cdot \arcsin(x) + \frac{1}{\sqrt{1 – x^2}}\right);
\]
So, the final result is \( f'(x) = e^{2 \cdot x} \cdot \left(2 \cdot \arcsin(x) + \frac{1}{\sqrt{1 – x^2}}\right) \).
It Gets Even More Interesting: Where to Go After the Derivative of Arcsin
You have already understood the derivative of arcsin, so the logical next step is to move on to other inverse trigonometric functions. This way, you will see the common ideas in the proofs, learn to quickly spot similar techniques, and feel more confident working with more complex expressions. Below are the topics that are worth studying next.
- Derivative of Arccosine: Formula, Proof, Examples — You will see how it is connected to arcsin, get a step-by-step proof, and practice with problems ranging from basic to more combined ones.
- Derivative of Arctangent: Formula, Proof, Examples — You will learn how the derivative of arctangent behaves, and how to apply the formula and the chain rule in problems with different algebraic transformations.
- Derivative of Arccotangent: Formula, Proof, Examples — You will examine the formula and proof of the derivative of arccotangent, and also solve several examples to firmly reinforce the material.
If you are already solving problems on derivatives but sometimes doubt your result, use an online derivative calculator to quickly check your work.
Derivative of Arcsin in Programming: From Flowchart to Code
If you enjoy programming and want to bring the theory to life in practice, try implementing a ready-made flowchart of an algorithm that finds the equation of the tangent line to the graph of the arcsin(x) function at a point chosen by the user. Step by step, turn each block of the flowchart into real programming instructions, add a convenient way to enter the data, and make the output of the result clear and user-friendly.
This kind of task not only improves your logical thinking and coding skills, but also shows how the derivative of arcsin becomes a practical tool for solving applied problems. Try to implement the algorithm on your own — and the flowchart will turn into a working program.
