**The derivative of the inverse cosine function is equal to minus 1 over the square root of 1 minus x squared, -1/(√(1-x ^{2})).** 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.

##### CALCULUS

**Relevant for**…

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

## Avoid confusion in using the denotations arccos(x), cos^{-1}(x),
1
/
cos(x) , and cos^{n}(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 $latex f(x) = \cos^{-1}(8x)$?

##### Solution

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 $latex F(x) = \cos^{-1}(x^2+10)$

##### Solution

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

##### Solution

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

## See also

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

- Derivative of arcsin (Inverse Sine) 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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