**The derivative of the secant function is secant times tangent, sec(x)tan(x)**. We can prove this derivative using limits and trigonometric identities.

In this article, we will learn how to derive the trigonometric secant function. We will cover a demonstration of the derivative, a comparison of graphs of the secant and its derivative, and some examples.

## Proof of the Derivative of the Secant Function

The trigonometric function *secant* of an angle is defined as the ratio of hypothenuse to the adjacent side of an angle in a right triangle. Illustrating it through a figure, we have

where C is 90°. For the sample right triangle, getting the secant of angle *A* can be evaluated as

$latex \sec{(A)} = \frac{c}{b}$

where *A* is the angle, *c* is the hypothenuse, and *b* is its adjacent side.

Before learning the proof of the derivative of the secant function, you are hereby recommended to learn the Pythagorean theorem, Soh-Cah-Toa & Cho-Sha-Cao, and the first principle of limits as prerequisites.

To review, any function can be derived by equating it to the limit of

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

Suppose we are asked to get the derivative of

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

we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sec{(x+h)} – \sec{(x)} }{h}}$$

Analyzing our equation, we can observe that both the first and second terms in the numerator of the limit is a secant of a sum of two angles *x* and *h* and a secant of angle *x*. With this observation, we can try to apply the *defining relation identities for secant and cosine*. Applying this, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \frac{1}{\cos{(x+h)}} – \frac{1}{\cos{(x)}} }{h}}$$

Algebraically re-arranging by applying some rules of fraction, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \frac{ \cos{(x)} – \cos{(x+h)} }{\cos{(x+h)}\cos{(x)}} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \frac{ \cos{(x)} – \cos{(x+h)} }{h\cos{(x+h)}\cos{(x)}} }$$

Looking at the re-arranged numerator, we can try to apply the *product-sum identities of cosine*.

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -2\sin{\left(\frac{x+(x+h)}{2}\right)} \sin{\left(\frac{x-(x+h)}{2}\right)} }{h\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -2\sin{\left(\frac{x+x+h}{2}\right)} \sin{\left(\frac{x-x-h}{2}\right)} }{h\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -2\sin{\left(\frac{2x+h}{2}\right)} \sin{\left(\frac{-h}{2}\right)} }{h\cos{(x+h)}\cos{(x)}} \right)}$$

Based on the trigonometric identities of a sine of a negative angle, it is equal to negative sine of that same angle but in positive form. Applying this to the second multiplicand of the numerator, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -2\sin{\left(\frac{2x+h}{2}\right)} \cdot \left( -\sin{\left(\frac{h}{2}\right)} \right) }{h\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ 2\sin{\left(\frac{2x+h}{2}\right)} \sin{\left(\frac{h}{2}\right)} }{h\cos{(x+h)}\cos{(x)}} \right)}$$

Re-arranging algebraically and by applying the limit of product of two functions, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ \sin{\left(\frac{2x+h}{2}\right)} \cdot 2\sin{\left(\frac{h}{2}\right)} }{\cos{(x+h)}\cos{(x)} \cdot h} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+h}{2}\right)} }{ \cos{(x+h)}\cos{(x)} } \cdot \frac{ 2\sin{\left(\frac{h}{2}\right)} }{h} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+h}{2}\right)} }{ \cos{(x+h)}\cos{(x)} } \right)} \cdot \lim \limits_{h \to 0} {\left( \frac{ 2\sin{\left(\frac{h}{2}\right)} }{h} \right)}$$

Applying some rules of fractions to the second multiplicand, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+h}{2}\right)} }{ \cos{(x+h)}\cos{(x)} } \right)} \cdot \lim \limits_{h \to 0} {\left( \frac{ \sin{\left(\frac{h}{2}\right)} }{ \frac{h}{2} } \right)}$$

In accordance with the limits of trigonometric functions, the limit of the trigonometric function $latex \sin{(\theta)}$ to $latex \theta$ as $latex \theta$ approaches zero is equal to one. The same can be applied to $latex \sin{\left(\frac{h}{2}\right)}$ over $latex \frac{h}{2}$. Applying, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+h}{2}\right)} }{ \cos{(x+h)}\cos{(x)} } \right)} \cdot \lim \limits_{h \to 0} {1}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+h}{2}\right)} }{ \cos{(x+h)}\cos{(x)} } \right)}$$

Finally, we have successfully made it possible to evaluate the limit of whatever is left in the equation. Evaluating by substituting the approaching value of $latex h$, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x+(0)}{2}\right)} }{ \cos{(x+(0))}\cos{(x)} } \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{\sin{\left(\frac{2x}{2}\right)} }{ \cos{(x)}\cos{(x)} } \right)}$$

$$\frac{d}{dx} f(x) = \frac{\sin{(x)} }{ \cos{(x)}\cos{(x)} }$$

Applying some trigonometric identities to simplify the derivative formula by the use of *defining relation identities*, we have

$$\frac{d}{dx} f(x) = \frac{ \sin{(x)} }{ \cos{(x)} } \cdot \frac{1}{\cos{(x)}} $$

$$\frac{d}{dx} f(x) = \tan{(x)} \cdot \sec{(x)} $$

$$\frac{d}{dx} f(x) = \sec{(x)} \tan{(x)}$$

Therefore, the derivative of the trigonometric function ‘*secant*‘ is:

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

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

The graph of the function

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

is

Deriving $latex f(x) = \sec{(x)}$, we get

$latex f'(x) = \sec{(x)}\tan{(x)}$

which is illustrated graphically as

Making a comparison of their graphs, we have

We can observe that the original function $latex f(x) = \sec{(x)}$ has a domain of

$latex \left(-\frac{3\pi}{2},-\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$

*within the finite intervals of*

$latex \left(-\frac{3\pi}{2},\frac{3\pi}{2}\right)$

and exists within the range of

$latex (-\infty,-1] \cup [1,\infty)$

whereas the derivative $latex f'(x) = \sec{(x)}\tan{(x)}$ has a domain of

$latex \left(-\frac{3\pi}{2},-\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$

*within the finite intervals of*

$latex \left(-\frac{3\pi}{2},\frac{3\pi}{2}\right)$

and exists within the range of

$latex (-\infty,\infty)$ or *all real numbers*

## Examples

The following are some examples of how to derive compound secant functions with the chain rule.

### EXAMPLE 1

Find the derivative of $latex f(x) = \sec(10x)$.

##### Solution

The given secant function is a composite function since it is a secant of the function $latex 10x$. Therefore, we are going to use the chain rule.

If $latex u=10x$ is the inner function, we have $latex f(u)=\sec(u)$ and using the chain rule, we have:

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

$$\frac{dy}{dx}=\sec(u)\tan(u) \times 10$$

Finally, we plug $latex u=10x$ back into the function and we have:

$$\frac{dy}{dx}=10\sec(10x)\tan(10x)$$

### EXAMPLE 2

Find the derivative of $latex F(x) = \sec(8x^2-4)$

##### Solution

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

We start by expressing the secant function as $latex f (u) = \sec(u)$, where $latex u = 8x^2-4$.

Then, we calculate the derivative of the exterior function $latex f(u)$, that is, the derivative of the secant function in terms of $latex u$:

$$\frac{d}{du} ( \sec{(u)} ) = \sec(u)\tan(u)$$

Now, we calculate the derivative of the inner function $latex g(x)$ or $latex u$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(8x^2-4)$$

$$\frac{d}{dx}(g(x)) = 16x$$

Using the chain rule, we multiply the derivative of the outer function by the derivative of the inner function:

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

$$\frac{dy}{dx} = \sec(u)\tan(u) \cdot 16x$$

Finally, we substitute $latex u=8x^2-4$ back:

$$\frac{dy}{dx} = \sec(8x^2-4)\tan(8x^2-4) \cdot 16x$$

$$\frac{dy}{dx} = 16x\sec(8x^2-4)\tan(8x^2-4)$$

### EXAMPLE 3

What is the derivative of $latex f(x) = \sec(\sqrt{x})$?

##### Solution

We can derive this function by considering $latex u=\sqrt{x}$ as the inner function.

Therefore, we start by writing $latex u=\sqrt{x}$ as $latex u=x^{\frac{1}{2}}$ to find the derivative of $latex u$ in terms of $latex x$:

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

Now, we consider that $latex f(u)=\sec(u)$ and use the chain rule:

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

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

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

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

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

## Practice of derivatives of composite secant functions

## See also

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