The squared of a secant function is the secant function raised to the power of two. **The derivative of secant squared is equal to two times tangent times secant squared, 2tan(x)sec ^{2}(x).** This derivative can be found by using the chain rule and the derivatives of the fundamental trigonometric functions.

We’ll go through the basics, definition, formula, graph comparison of underived and derived secant squared, proof, derivation techniques, and some more relevant instances.

## Proof of The Derivative of Secant Squared Function

As a prerequisite, please review the chain rule formula and its proof by looking here: Chain Rule of derivatives. Similarly, you can review the proof of the derivative of secant function by visiting this link: Derivative of Secant, sec(x).

Please be reminded that

$latex \sec^{2}{(x)} \neq \sec{(x^2)}$

To prevent misunderstanding, the first is a “full trigonometric function” raised to the power of two, whereas the second is a trigonometric function of “the square of a variable.”

Because it is a composite function, the chain rule formula is used as a more straightforward tool to prove the derivative of the secant squared function, providing you already know how to prove the chain rule formula and the derivative of a secant function.

Assuming we are asked to find the derivative of

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

We can identify the two functions that make up F(x). There is a power function and a trigonometric function in this scenario. To be more exact, these are a function raised to a power of two and a trigonometric function of secant, based on our given F(x).

For a better representation, we can rewrite it as

$latex \frac{dy}{dx} = \sec^{2}{(x)}$

$latex \frac{dy}{dx} = (\sec{(x)})^2$

It becomes evident that the given power function is the outer function to be considered, while the secant function, being raised by the given power function, is the inner function. We can configure the outer function as follows:

$latex f(u) = u^2$

where

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

The trigonometric secant function, as the inner function of *f(u)*, will be denoted as *g(x)*.

$latex f(u) = f(g(x))$

$latex u = g(x)$

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

Deriving the outer function *f(u)* using the power rule in terms of *u*, we have

$latex f(u) = u^2$

$latex f'(u) = 2u$

Deriving the inner function *g(x)* using the derivative formula of trigonometric function *secant* in terms of *x*, we have

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

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

Algebraically multiplying the derivative of outer function $latex f'(u)$ by the derivative of inner function $latex g'(x)$, we have

$latex \frac{dy}{dx} = f'(u) \cdot g'(x)$

$latex \frac{dy}{dx} = (2u) \cdot (\sec{(x)}\tan{(x)})$

Substituting *u* into *f'(u)*, we have

$latex \frac{dy}{dx} = (2(\sec{(x)})) \cdot (\sec{(x)}\tan{(x)})$

$latex \frac{dy}{dx} = 2\sec^{2}{(x)} \cdot \tan{(x)}$

which brings us to the derivative formula of secant squared *x*

$latex \frac{d}{dx} \sec^{2}{(x)} = 2\tan{(x)}\sec^{2}{(x)}$

## Relationship between the derivative of secant squared and tangent squared, what makes them similar?

You may wonder why

$latex \sec^{2}{(x)}$

and

$latex \tan^{2}{(x)}$

have similar derivatives.

According to the Pythagorean formula for secants and tangents,

$latex \sec^{2}{(x)} = 1 + \tan^{2}{(x)}$

If we try to derive both sides of the equation, we have

$latex \frac{d}{dx} (\sec^{2}{(x)}) = \frac{d}{dx}(1) + \frac{d}{dx}(\tan^{2}{(x)})$

Evaluating the derivative of the first term in the right-hand-side of the equation, where the derivative is zero, we have

$latex \frac{d}{dx} (\sec^{2}{(x)}) = 0 + \frac{d}{dx}(\tan^{2}{(x)})$

$latex \frac{d}{dx} (\sec^{2}{(x)}) = \frac{d}{dx}(\tan^{2}{(x)})$

This is why both the secant squared and the tangent squared have the same derivative, because of the Pythagorean formula for secants and tangents.

## How is a Secant Squared Function derived? *Quicker Methods*

Secant squared, as previously stated, is a composite function of power and the trigonometric function secant. Instead of using the chain rule formula repetitively like what we did in the proof, we may simply use the established derivative formula for a secant squared function.

### METHOD 1: Derivative of the square of a secant of any angle *x* in terms of the same angle *x*.

$latex \frac{d}{dx} \left( \sec^{2}{(x)} \right) = 2\tan{(x)}\sec^{2}{(x)}$ |

**Step 1:** Determine whether the secant squared of an angle is a function of the same angle. For instance, if the right-hand side of the equation is $latex \sec^{2}{(x)}$, determine if it is a function of the same angle *x* or *f(x)*.

* Note:* If $latex \sec^{2}{(x)}$ is a function of a different angle or variable, such as

*f(t)*or

*f(y)*, implicit differentiation will be used, which is beyond the scope of this article.

** Step 2: **Then directly apply the proven derivative formula of the secant squared function

$latex \frac{dy}{dx} = 2\tan{(x)}\sec^{2}{(x)}$

If nothing else can be simplified, then that is the final solution.

### METHOD 2: Derivative of the square of a secant of any function *v* in terms of *x.*

$latex \frac{d}{dx} \left( \sec^{2}{(v)} \right) = 2\tan{(v)}\sec^{2}{(v)} \cdot \frac{d}{dx} (v)$ |

** Step 1:** Express the function as $latex G(x) = \sec^{2}{(v)}$, where $latex v$ represents any function other than

*x*.

** Step 2:** Treat $latex \sec^{2}{(v)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. Doing this, we have

$latex g(v) = \sec{(v)}$

and

$latex h(x) = v$

** Step 3:** Derive the outer function $latex g(v)$, and use the derivative of the secant squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \sec^{2}{(v)} \right) = 2\tan{(v)}\sec^{2}{(v)}$

** Step 4:** Derive the inner function $latex h(x) = v$. Use whatever appropriate derivative rule applies to $latex v$.

** Step 5:** Algebraically multiply the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$ to completely apply chain rule

$latex \frac{dy}{dx} = \frac{d}{du} (g(v)) \cdot \frac{d}{dx} (h(x))$

$latex \frac{dy}{dx} = 2\tan{(v)}\sec^{2}{(v)} \cdot \frac{d}{dx} (v)$

** Step 6:** Substitute $latex v$ into $latex g'(v)$

** Step 7:** Simplify and apply any function law whenever applicable then finalize the answer.

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

With the function

$latex f(x) = \sec^{2}{(x)}$

it is graphed as

As we already know, deriving $latex f(x) = \sec^{2}{(x)}$ is

$latex f'(x) = 2\tan{(x)}\sec^{2}{(x)}$

which is graphed as

Illustrating both graphs in one, we have

By examining the differences between these functions basing on the graphs above, you can see that the original function $latex f(x) = \sec^{2}{(x)}$ 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 lies within the range of

$latex [1,\infty)$

whereas the derivative $latex f'(x) = 2\tan{(x)}\sec^{2}{(x)}$ 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 lies within the range of

$latex (-\infty,\infty)$

## Examples

Here are some examples of deriving a secant squared function using either the first or second method.

### EXAMPLE 1

**Derive:** $latex f(\beta) = \sec^{2}{(\beta)}$

**Solution:** After examining the given secant squared function, it shows that it is only a square of a secant of a single angle $latex \beta$. Hence, we can apply the first method to this problem.

**Step 1:** Assess if the square of secant of $latex \beta$ is a function of $latex \beta$. It is in this problem. Hence, proceed to step 2.

**Step 2: **Directly apply the derivative formula of the secant squared function and derive in terms of $latex \beta$. Since no further simplification is needed, **the final answer is:**

$latex f'(\beta) = 2\tan{(\beta)}\sec^{2}{(\beta)}$

### EXAMPLE 2

**Derive:** $latex G(x) = \sec^{2}{(3x^2-7)}$

**Solution:** After examining the given secant squared function, it shows that it is a square of a secant of a polynomial function. Hence, we can apply the second method to this problem.

**Step 1:** Express the secant squared function as $latex G(x) = \sec^{2}{(v)}$, using $latex v$ to represent any function other than *x*, which is the angle of the secant squared. In this problem,

$latex v = 3x^2-7$

Let’s substitute this later as we evaluate the derivative of the problem.

**Step 2:** Consider $latex \sec{(v)}$ as the outside function $latex g(v)$. Then $latex v$ will be the inner function, denoted as $latex h(x)$ too. For this problem, we have

$latex g(v) = \sec{(v)}$

and

$latex h(x) = v = 3x^2-7$

**Step 3:** Derive the outer function $latex g(v)$ using the derivative of the secant squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \sec^{2}{(v)} \right) = 2\tan{(v)}\sec^{2}{(v)}$

**Step 4:** Derive the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a polynomial function, let’s apply the power rule and sum/difference of derivatives to derive $latex v$.

$latex \frac{d}{dx}(h(x)) = \frac{d}{dx} \left(3x^2-7 \right)$

$latex \frac{d}{dx}(h(x)) = 6x$

**Step 5:** Evaluate the basic chain rule formula by algebraically multiplying the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (g(v)) \cdot \frac{d}{dx} (h(x))$

$latex \frac{dy}{dx} = 2\tan{(v)}\sec^{2}{(v)} \cdot 6x$

**Step 6:** Substitute $latex v$ into $latex g'(v)$

$$\frac{dy}{dx} = 2\tan{(v)}\sec^{2}{(v)} \cdot 6x$$

$$\frac{dy}{dx} = 2\tan{(3x^2-7)}\sec^{2}{(3x^2-7)} \cdot 6x$$

**Step 7:** Simplify and apply any function law whenever applicable to finalize the answer.

$$\frac{dy}{dx} = 2\cdot6x \cdot \tan{(3x^2-7)}\sec^{2}{(3x^2-7)}$$

$$\frac{dy}{dx} = 6x\tan{(3x^2-7)}\sec^{2}{(3x^2-7)}$$

And **the final answer is:**

$$G'(x) = 6x\tan{(3x^2-7)}\sec^{2}{(3x^2-7)}$$

## See also

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

- Derivative of Cosine Squared, cos^2(x) with Proof and Graphs
- Derivative of Sine Squared, sin^2(x) with Proof and Graphs
- Derivative of Tangent Squared, tan^2(x) with Proof and Graphs
- Derivative of Cosecant Squared, csc^2(x) with Proof and Graphs
- Derivative of Cotangent Squared, cot^2(x) with Proof and Graphs

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