# Derivative of arccos (Inverse Cosine) With Proof and Graphs

The derivative of the inverse cosine function is equal to minus 1 over the square root of 1 minus x squared, -1/(√(1-x2)). This derivative can be proved using the Pythagorean theorem and algebra.

In this article, we will learn how to derive the inverse cosine function. We’ll cover brief basics, a proof, a graphical comparison of the function and its derivative, and some examples.

##### CALCULUS

Relevant for

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

See proof

##### CALCULUS

Relevant for

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

See proof

## Avoid confusion in using the denotations arccos(x), cos-1(x), 1 / cos(x) , and cosn(x)

It is important to avoid possible confusion by using the different denotations $latex \arccos{(x)}$, $latex \cos^{-1}{(x)}$, $latex \frac {1}{\cos{ (x)}}$ and $latex \cos^{n}{(x)}$ because it could lead to derivation errors.

Summarizing the definition of these symbols, we have

$latex \arccos{(x)} = \cos^{-1}{(x)}$

The symbols $latex \arccos$ and $latex \cos^{-1}$ are used to represent inverse cosine. $latex \arccos$ is commonly used as the verbal symbol for the inverse cosine function, while $latex \cos^{-1}$ is used as a mathematical symbol for the inverse cosine function for a more formal function.

However, in the case of the denotation $latex \cos^{-1}{(x)}$, we must consider that $latex -1$ is not an algebraic exponent of a non-inverse cosine. The $latex -1$ used for the inverse cosine represents that the cosine is inverse and not raised to $latex -1$.

Therefore,

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

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

## Proof of the Derivative of the Inverse Cosine Function

In this proof, we will mainly use the concepts of a right triangle, the Pythagorean theorem, the trigonometric function of cosine and sine, 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 hypothenuse, there is a side x perpendicular to side $latex \sqrt{1-x^2}$ and an angle y adjacent to side x and opposite 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 cosine function by using its adjacent side x and the hypothenuse 1.

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

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

$latex cos{(y)} = x$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Graph of Inverse Cosine x VS. The Derivative of Inverse Cosine x

The graph of the function

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

is

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

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

which has the following graph

Comparing their graphs, we have

Using the graphs, we can see that the original function $latex f(x) = \cos^{-1}{(x)}$ has a domain of

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

and exists within the range of

$latex [0,\pi]$ or $latex 0 \leq y \leq \pi$

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

$latex (-\infty,-1]$ or $latex y \leq -1$

## Examples

The following examples show how to derive composite inverse cosine functions.

### EXAMPLE 1

What is the derivative of $latex f(x) = \cos^{-1}(8x)$?

We have a composite inverse cosine function. Therefore, we have to use the chain rule to derive it.

Considering that $latex u=8x$ is the inner function, we have $latex f(u)=\cos^{-1}(u)$. Then, applying the chain rule, we have:

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

$$\frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}} \times 8$$

Finally, we substitute $latex u=8x$ back into the function and we have:

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

$$\frac{dy}{dx}=-\frac{8}{\sqrt{1-64x^2}}$$

### EXAMPLE 2

Find the derivative of the function $latex F(x) = \cos^{-1}(x^2+10)$

Let’s use the chain rule by writing the inverse cosine function as $latex f (u) = \cos^{-1}(u)$, where $latex u = x^2+10$.

Therefore, we start by writing the derivative of the outer function $latex f(u)$:

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

Now, we calculate the derivative of the inner function $latex g(x)=u=x^2+10$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^2+5)$$

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

Then, we multiply the derivative of the external function by the derivative of the internal function and we have:

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

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

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

$$\frac{dy}{dx} = -\frac{1}{\sqrt{1-(x^2+10)^2}} \cdot 2x$$

$$\frac{dy}{dx} = -\frac{2x}{\sqrt{1-(x^4+20x^2+100)}}$$

$$\frac{dy}{dx} = -\frac{2x}{\sqrt{-x^4-20x^2-99)}}$$

### EXAMPLE 3

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

In this case, the square root function is the inner function. We can derive it by writing $latex u=\sqrt{x}$ as $latex u=x^{\frac{1}{2}}$:

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

Now, considering that $latex f(u)=\cos^{-1}(u)$, we apply 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 cosine functions

Derivatives of inverse cosine quiz  You have completed the quiz!

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

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  