**The derivative of the inverse sine function is equal to 1 over square root of 1 minus x squared, 1/(√(1-x ^{2}))**. We can prove this derivative using the Pythagorean theorem and algebra.

In this article, we will learn how to derive the inverse sine function. We will look at a proof, a graphical comparison of the function and its derivative, and will explore some examples.

##### CALCULUS

**Relevant for**…

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

##### CALCULUS

**Relevant for**…

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

## Avoid confusion in using the denotations arcsin(x), sin^{-1}(x),
1
/
sin(x) , and sin^{n}(x)

It is important to avoid possible confusions that we could have when using denotations $latex \arcsin{(x)}$, $latex \sin^{-1}{(x)}$, $latex \frac {1}{\sin{( x)}}$ and $latex \sin^{n}{(x)}$, since interchanging the meaning of these symbols can lead to derivation errors.

Summarizing the definition of these symbols, we have

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

Both the symbols $latex \arcsin$ and $latex \sin^{-1}$ can be used when calculating the inverse sine of a variable or other function. $latex \arcsin$ is commonly used as the verbal symbol for the inverse sine function, while $latex \sin^{-1}$ is used as a mathematical symbol for the inverse sine function for a more formal function.

In the case of the denotation $latex \sin^{-1}{(x)}$, it can sometimes confuse students that $latex -1$ is an algebraic exponent of a non-inverse sine, which is not true. The $latex -1$ used for the inverse sine represents that the sine is inverse and not raised to $latex -1$.

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.

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*.

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

The graph of the function

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

is

And when deriving $latex f(x) = \sin^{-1}{(x)}$, we get

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

which has the following graph

Comparing both graphs in one, we have

Using the graph, we can see 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

The following examples show how to derive a composite inverse sine function.

### EXAMPLE 1

What is the derivative of $latex f(x) = \sin^{-1}(7x)$?

##### Solution

We can derive this function using the chain rule by considering that we have an inverse sine of $latex 7x$.

Therefore, we can consider $latex u=7x$ as the inner function, and we have $latex f(u)=\sin^{-1}(u)$. Applying the chain rule, we have:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}} \times 7$$

Finally, we substitute $latex u=7x$ back into the function and we have:

$$\frac{dy}{dx}=\frac{7}{\sqrt{1-(7x)^2}}$$

$$\frac{dy}{dx}=\frac{7}{\sqrt{1-49x^2}}$$

### EXAMPLE 2

Derive the function $latex F(x) = \sin^{-1}(x^3+5 )$

##### Solution

We are going to use the chain rule. Therefore, we write the inverse sine function as $latex f (u) = \sin^{-1}(u)$, where $latex u = x^3+5$.

Then, we find the derivative of the outer function $latex f(u)$:

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

Now, we find the derivative of the inner function $latex g(x)=u=x^3+5$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^3+5)$$

$$\frac{d}{dx}(g(x)) = 3x^2$$

Multiplying the derivative of the outer function by the derivative of the inner function, we have:

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

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot 3x^2$$

Finally, we substitute $latex u$ back in and simplify:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-(x^3+5)^2}} \cdot 3x^2$$

$$\frac{dy}{dx} = \frac{3x^2}{1-(x^6+10x^3+25)}$$

$$F'(x) = \frac{3x^2}{-x^6-10x^3-24}$$

### EXAMPLE 3

Find the derivative of $latex f(x) = \sin^{-1}(\sqrt{x})$

##### Solution

The square root is the inner function in this case. Therefore, we consider that $latex u=\sqrt{x}$ is equal to $latex u=x^{\frac{1}{2}}$ and find its derivative:

$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$$

Now, we write $latex f(u)=\tan(u)$ and use the chain rule:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

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

Substituting $latex u=\sqrt{x}$ and simplifying, we have:

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

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

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

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

## Practice of derivatives of composite inverse sine functions

## See also

Interested in learning more about the derivatives of trigonometric functions? Take a look at these pages:

- Derivative of arccos (Inverse Cosine) With Proof and Graphs
- Derivative of arctan (Inverse Tangent) With Proof and Graphs
- Derivative of arcsec (Inverse Secant) With Proof and Graphs
- Derivative of arccsc (Inverse Cosecant) With Proof and Graphs
- Derivative of arccot (Inverse Cotangent) With Proof and Graphs

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