Derivative of arcsin (Inverse Sine) With Proof and Graphs

The Derivative of ArcSine or Inverse Sine is used in deriving a function that involves the inverse form of the trigonometric function ‘sine‘. The derivative of the inverse sine function is equal to 1 over square root of 1 minus x squared, 1/(√(1-x2)). We can prove this derivative using the Pythagorean theorem and algebra.

In this article, we will discuss how to derive the arcsine or the inverse sine function. We will cover brief fundamentals, its definition, formula, a graph comparison of the underived and derived function, a proof, methods to derive, and a few examples.

CALCULUS
Derivative of arcsine, inverse sine

Relevant for

Learning about the proof and graphs of the derivative of arsin of x.

See proof

CALCULUS
Derivative of arcsine, inverse sine

Relevant for

Learning about the proof and graphs of the derivative of arsin of x.

See proof

Avoid confusion in using the denotations arcsin(x), sin-1(x),
1 / sin(x)
, and sinn(x)

It is important that we do not compromise the potential confusions we might have in using different denotations between $latex \arcsin{(x)}$, $latex \sin^{-1}{(x)}$, $latex \frac{1}{\sin{(x)}}$, and $latex \sin^{n}{(x)}$, since interchanging the meaning of these symbols may lead to derivation mistakes. Summarizing the definition of these symbols, we have

$latex \arcsin{(x)} = \sin^{-1}{(x)}$

Both symbols $latex \arcsin$ and $latex \sin^{-1}$ can be used interchangeably when computing for the inverse sine of either a variable or another function. $latex \arcsin$ is commonly used as the verbal symbol of inverse sine function which is popularly used as introductory denotations for beginners whilst $latex \sin^{-1}$ is used as a mathematical symbol of inverse sine function for a more professional setting.

However, when it comes to the denotation $latex \sin^{-1}{(x)}$, sometimes it can confuse learners that $latex -1$ is an algebraic exponent of a non-inverse sine, which is not true. The $latex -1$ used for inverse sine represents the sine being inverse and not raised to $latex -1$. This has been proven and shown in the previous sub-article written above.

Therefore,

$latex \sin^{-1}{(x)} \neq \frac{1}{\sin{(x)}}$

And givens such as $latex \sin^{2}{(x)}$ or $latex \sin^{n}{(x)}$, where n is any algebraic exponent of a non-inverse sine, MUST NOT use the inverse sine formula since in these givens, both the 2 and any exponent n are treated as algebraic exponents of a non-inverse sine.


Proof of the Derivative of the Inverse Sine Function

In this proof, we will mainly use the concepts of a right triangle, the Pythagorean theorem, the trigonometric function of sine and cosine, and some basic algebra. Just like in the previous figure as a reference sample for a given right triangle, suppose we have that same triangle $latex \Delta ABC$, but this time, let’s change the variables for an easier illustration.

Right-Triangle-siny-fracsqrt1-x2hyp

where for every one-unit of the hypotenuse, there is a side $latex \sqrt{1-x^2}$ perpendicular to the side x and an angle y opposite to side x and adjacent to $latex \sqrt{1-x^2}$.

Using these components of a right-triangle, we can find the angle y by using Soh-Cah-Toa, particularly the sine function by using its opposite side x and the hypothenuse 1.

$latex \sin{(\theta)} = \frac{opp}{hyp}$

$latex \sin{(y)} = \frac{x}{1}$

$latex \sin{(y)} = x$

Now, we can implicitly derive this equation by using the derivative of trigonometric function of sine for the left-hand side and power rule for the right-hand side. Doing so, we have

$latex \frac{d}{dx} (\sin{(y)}) = \frac{d}{dx} (x)$

$latex \frac{d}{dx} (\cos{(y)}) = 1$

$latex \frac{dy}{dx} (\cos{(y)}) = 1$

$latex \frac{dy}{dx} = \frac{1}{cos{(y)}}$

Getting the cosine of our given right-triangle, we have

$latex \cos{(y)} = \frac{adj}{hyp}$

$latex \cos{(y)} = \frac{\sqrt{1-x^2}}{1}$

$latex \cos{(y)} = \sqrt{1-x^2}$

We can then substitute $latex \cos{(y)}$ to the implicit differentiation of $latex \sin{(y)} = x$

$latex \frac{dy}{dx} = \frac{1}{cos{(y)}}$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$

Therefore, algebraically solving for the angle y and getting its derivative, we have

$latex \sin{(y)} = x$

$latex y = \frac{x}{\sin}$

$latex y = \sin^{-1}{(x)}$

$latex \frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1}{(x)} \right)$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$

which is now the derivative formula for the inverse sine of x.

Now, for the derivative of an inverse sine of any function other than x, we may apply the derivative formula of inverse sine together with the chain rule formula. By doing so, we have

$latex \frac{dy}{dx} = \frac{d}{du} \sin^{-1}{(u)} \cdot \frac{d}{dx} (u)$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{d}{dx} (u)$

where $latex u$ is any function other than x.


How to derive an Inverse Sine Function?

The derivative process of an inverse sine function is very straightforward assuming you have already learned the concepts behind the usage of the inverse sine function and how we arrived to its derivative formula.

METHOD 1: Derivative of the Inverse Sine of any single variable x

$latex \frac{d}{dx} \left( \sin^{-1}{(x)} \right) = \frac{1}{\sqrt{1-x^2}}$

Step 1: Analyze if the inverse sine of a variable is a function of that same variable. For example, if the right-hand side of the equation is $latex \sin^{-1}{(x)}$, then check if it is a function of the same variable x or f(x). After this, proceed to Step 2 until you complete the derivation steps.

Note: If $latex \sin^{-1}{(x)}$ is a function of a different variable such as f(t) or f(y), it will use implicit differentiation which is out of the scope of this article.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

$latex \frac{d}{dx} f(x) = \frac{d}{dx} \left(\sin^{-1}{(x)} \right)$

or

$latex \frac{dy}{dx} = \frac{d}{dx} \left(\sin^{-1}{(x)} \right)$

Step 3: Then directly apply the derivative formula of inverse sine function

$latex \frac{dy}{dx} = \frac{1}{\sqrt{(1-x^2)}}$

If nothing is to be simplified anymore, then that would be the final answer.

METHOD 2: Derivative of the Inverse Sine of any function u in terms of x.

$latex \frac{d}{dx} \left( \sin^{-1}{(u)} \right) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{d}{dx} (u)$

Step 1: Express the inverse function as $latex F(x) = \sin^{-1}{(u)}$ or $latex F(x) = \arcsin{(u)}$, where $latex u$ represents any function other than x.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

$latex \frac{d}{dx} F(x) = \frac{d}{dx} \left(\sin^{-1}{(u)} \right)$

or

$latex \frac{dy}{dx} = \frac{d}{dx} \left(\sin^{-1}{(u)} \right)$

Step 3: Consider $latex \sin^{-1}{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. Hence we have

$latex f(u) = \sin^{-1}{(u)}$

and

$latex g(x) = u$

Step 4: Get the derivative of the outer function $latex f(u)$, which must use the derivative of inverse sine $latex \sin^{-1}{(u)}$, in terms of $latex u$.

$latex \frac{d}{du} \left( \sin^{-1}{(u)} \right) = \frac{1}{\sqrt{1-u^2}}$

Step 5: Get the derivative of the inner function $latex g(x) = u$. Use the appropriate derivative rule that applies to $latex u$.

Step 6: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex f(u)$ by the derivative of inner function $latex g(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{d}{dx} (u)$

Step 7: Substitute $latex u$ into $latex f'(u)$

Step 8: Simplify and apply any function law whenever applicable to finalize the answer.


Graph of Inverse Sine x VS. The Derivative of Inverse Sine x

Given the function

$latex f(x) = \sin^{-1}{(x)}$

the graph is illustrated as

graph-of-arcsinx

And as we know by now, by deriving $latex f(x) = \sin^{-1}{(x)}$, we get

$latex f'(x) = \frac{1}{\sqrt{1-x^2}}$

which is illustrated graphically as

graph of derivative of arcsin x

Illustrating both graphs in one, we have

graph of arcsin x and its derivative

Analyzing the differences of these functions through these graphs, you can observe that the original function $latex f(x) = \sin^{-1}{(x)}$ has a domain of

$latex [-1,1]$ or $latex -1 \leq x \leq 1$

and exists within the range of

$latex \left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$ or $latex -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

whereas the derivative $latex f'(x) = \frac{1}{\sqrt{1-x^2}}$ has a domain of

$latex (-1,1)$ or $latex -1 < x < 1$

and exists within the range of

$latex [1, \infty]$ or $latex y \geq 1$


Examples

Below are some examples of using either the first or second method in deriving an inverse sine function.

EXAMPLE 1

Derive: $latex f(\theta) = \sin^{-1}{(\theta)}$

Solution: Analyzing the given inverse sine function, it is only an inverse sine of a single variable $latex \theta$ raised to a variable equal to one. Therefore, we can use the first method to derive this problem.

Step 1: Analyze if the inverse sine of $latex \theta$ is a function of $latex \theta$. In this problem, it is. Hence, proceed to step 2.

Step 2: Illustrate derivation through

$latex \frac{d}{d\theta} f(\theta) = \frac{d}{d\theta} \left(\sin^{-1}{(\theta)} \right)$

Step 3: Directly apply the derivative formula of inverse sine function and derive in terms of $latex \theta$. Since no further simplification is needed, the final answer is:

$latex f'(\theta) = \frac{1}{\sqrt{1 – \theta^2}}$

EXAMPLE 2

Derive: $latex F(x) = \sin^{-1}{\left(4x^2+8 \right)}$

Solution: Analyzing the given inverse sine function, it is an inverse sine of a polynomial function. Therefore, we can use the second method to derive this problem.

Step 1: Express the inverse sine function as $latex F(x) = \sin^{-1}{(u)}$ or $latex F(x) = \arcsin{(u)}$, where $latex u$ represents any function other than x. In this problem,

$latex u = 4x^2+8$

We will substitute this later as we finalize the derivative of the problem.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

$latex \frac{d}{dx} F(x) = \frac{d}{dx} \left(\sin^{-1}{(u)} \right)$

Step 3: Consider $latex \sin^{-1}{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. For this problem, we have

$latex f(u) = \sin^{-1}{(u)}$

and

$latex g(x) = u = 4x^2+8$

Step 4: Get the derivative of the outer function $latex f(u)$, which must use the derivative of inverse sine $latex \sin^{-1}{(u)}$, in terms of $latex u$.

$latex \frac{d}{du} \left( \sin^{-1}{(u)} \right) = \frac{1}{\sqrt{1-u^2}}$

Step 5: Get the derivative of the inner function $latex g(x)$ or $latex u$. Since our $latex u$ in this problem is a polynomial function, we will use power rule and sum/difference of derivatives to derive $latex u$.

$latex \frac{d}{dx}(g(x)) = \frac{d}{dx} \left(4x^2+8 \right)$

$latex \frac{d}{dx}(g(x)) = 8x$

Step 6: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex f(u)$ by the derivative of inner function $latex g(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot 8x$

Step 7: Substitute $latex u$ into $latex f'(u)$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot 8x$

$latex \frac{dy}{dx} = \frac{1}{\sqrt{1-(4x^2+8)^2}} \cdot 8x$

Step 8: Simplify and apply any function law whenever applicable to finalize the answer.

$latex F'(x) = \frac{1}{\sqrt{1-(4x^2+8)^2}} \cdot 8x$

$latex F'(x) = \frac{8x}{\sqrt{1-(4x^2+8)^2}}$

And the final answer is:

$latex F'(x) = \frac{8x}{\sqrt{1-(4x^2+8)^2}}$


See also

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