# Derivative of arcsec (Inverse Secant) With Proof and Graphs

The derivative of the inverse secant function is equal to 1/(|x|√(x2-1)). We can prove this derivative using the Pythagorean theorem and algebra.

In this article, we will learn how to derive the inverse secant function. We will look at some basics, 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 arcsec of x.

See proof

##### CALCULUS

Relevant for

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

See proof

## Avoid confusion in using the denotations arcsec(x), sec-1(x), 1 / sec(x) , and secn(x)

It is important that we do not fall into the possible confusion that we may have when using different denotations $latex \text{arcsec}(x)$, $latex \sec^{-1}{(x)}$, $latex \frac{1 }{\sec{(x)}}$ and $latex \sec^{n}{(x)}$, as it can lead to derivation errors. Summarizing the definition of these symbols, we have

$latex \text{arcsec}(x) = \sec^{-1}{(x)}$

The symbols $latex \text{arcsec}$ and $latex \sec^{-1}$ are used interchangeably when calculating the inverse secant. $latex \text{arcsecant}$ is commonly used as the verbal symbol for the inverse secant function, while $latex \sec^{-1}$ is used as the mathematical symbol for the inverse secant function for a more formal setting.

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

Therefore,

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

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

## Proof of the Derivative of the Inverse Secant Function

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

Using these components of a right-triangle, we can find the angle y by using Cho-Sha-Cao, particularly the secant function by using the hypothenuse x and its adjacent side.

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

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

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

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

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

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

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

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

Getting the tangent of angle y from our given right-triangle, we have

$latex \tan{(y)} = \frac{opp}{adj}$

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

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

We can then substitute $latex \sec{(y)}$ and $latex \tan{(y)}$ to the implicit differentiation of $latex \sec{(y)} = x$

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

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

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

Now, since

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

and

$latex hypothenuse = x$

We know that a negative hypotenuse cannot exist. Therefore, $latex \sec{(y)}$ in this case cannot be negative. That’s why the x multiplicand in the denominator of the derivative of the inverse secant must be considered an absolute value.

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

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

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

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

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

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

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

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

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

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

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

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

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

The graph of the function

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

is

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

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

which has the following graph

Illustrating both graphs in one, we have

Using the graphs, it can be seen that the original function $latex f(x) = \sec^{-1}{(x)}$ has a domain of

$latex (-\infty,-1] \cup [1,\infty )$ or all real numbers except $latex -1 < x < 1$

and exists within the range of

$latex [0,\frac{\pi}{2}\big) \cup \big(\frac{\pi}{2},\pi]$ or $latex 0 \leq y \leq \pi$ except $latex \frac{\pi}{2}$

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

$latex (-\infty,-1) \cup (1,\infty)$ or all real numbers except $latex -1 \leq x \leq 1$

and exists within the range of

$latex (0,\infty)$ or $latex y > 0$

## Examples

In the following examples, we will see how to derive compound inverse secant functions.

### EXAMPLE 1

What is the derivative of $latex f(x) = \sec^{-1}(2x)$?

Since we have a composite inverse secant function, we can use the chain rule to derive it.

Therefore, we consider $latex u=2x$ as the inner function, and we have $latex f(u)=\sec^{-1}(u)$ and applying the chain rule, we have:

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

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

Then, we substitute $latex u=2x$ back into the function and we have:

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

$$\frac{dy}{dx}=\frac{2}{|2x|\sqrt{4x^2-1}}$$

### EXAMPLE 2

Find the derivative of the function $latex F(x) = \sec^{-1}(x^2-5)$

To use the chain rule, we write the inverse secant function as $latex f (u) = \sec^{-1}(u)$, where $latex u = x^2-5$.

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

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

Now, we calculate the derivative of the inner function $latex g(x)=u=x^2-5$:

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

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

Then, we have to multiply the derivative of the outer function by the derivative of the inner function:

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

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

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

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

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

$$F'(x) = \frac{2x}{|x^2-5|\sqrt{x^4-10x^2+24}}$$

### EXAMPLE 3

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

In this case, the inner function is $latex u=\sqrt{x}$. Considering that we can write it as $latex u=x^{\frac{1}{2}}$, its derivative is:

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

If we apply the chain rule with $latex f(u)=\sec^{-1}(u)$, we have:

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

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

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

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

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

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

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

## Practice of derivatives of compound inverse secant inverse functions

Derivatives inverse secant quiz
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## See also

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.

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