**The derivative of the inverse secant function is equal to 1/(|x|√(x ^{2}-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.

##### CALCULUS

**Relevant for**…

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

## Avoid confusion in using the denotations arcsec(x), sec^{-1}(x),
1
/
sec(x) , and sec^{n}(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 $latex f(x) = \sec^{-1}(2x)$?

##### Solution

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 $latex F(x) = \sec^{-1}(x^2-5)$

##### Solution

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

##### Solution

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

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