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

In this article, we’ll look at how to calculate the composite function tangent squared. We will go over the principles, definition, formula, graph comparison of underived and derived *tangent x squared*, proof, techniques to derive, and a few examples.

## Proof of The Derivative of Tangent Squared Function Using Chain Rule

You are encouraged to review the chain rule formula, as a prerequisite of this topic, by visiting this link: Chain Rule of derivatives. Likewise, you may visit this another link for the proof of the derivative of tangent function: Derivative of Tangent, tan(x).

Please take note that

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

Setting confusion aside, the former is a “whole trigonometric function” raised to the power of two, whereas the latter is a trigonometric function of “a variable raised to the power of two.”

Because it is a composite function, the chain rule formula is used to make the derivative formula of tangent squared function, provided you have already mastered the chain rule formula and the derivative of tangent function.

Suppose we are asked to get the derivative of

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

We can identify the two functions that make up *F(x)*. There is a power function and a trigonometric function in this scenario. Based on our given *F(x)*, they are a function raised to a power of two and a tangent trigonometric function.

For easier representation, we can rewrite our given as

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

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

It is evident now that the given power function is the outer function, while the tangent function squared by the given power function is the inner function. We can set the outer function as

$latex f(u) = u^2$

where

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

Setting the trigonometric tangent function as the inner function of *f(u)* by denoting it as *g(x)*, we have

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

$latex u = g(x)$

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

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 *tangent* in terms of *x*, we have

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

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

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

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

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

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

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

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

This gets us to the tangent squared *x* derivative formula.

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

## Relationship between the derivative of tangent squared and secant squared, why are they the same?

You may wonder why the derivative of both functions

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

and

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

are the same.

According to the Pythagorean formula for tangents and secants,

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

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, which is the derivative of a constant 1, 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 tangent squared and the secant squared have the same derivative.

## How to derive a Tangent Squared Function?

As noted previously, tangent squared is a composite function of power and the trigonometric function tangent. Instead of constantly using the chain rule method, we may simply utilize the established derivative formula for a tangent squared function.

### METHOD 1: When the square of a tangent of any angle *x* is to be derived in terms of the same angle *x*.

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

**Step 1:** Analyze if the tangent squared of an angle is a function of that same angle. For example, if the right-hand side of the equation is $latex \tan^{2}{(x)}$, then check if it is a function of the same angle *x* or *f(x)*.

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

*f(t)*or

*f(y)*, it will use implicit differentiation which is out of the scope of this article.

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

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

If nothing is to be simplified anymore, then that would be the final answer.

### METHOD 2: When the given is a tangent squared of any function *v* instead and to be derived in terms of *x.*

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

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

*x*.

** Step 2:** Consider $latex \tan^{2}{(v)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. Hence we have

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

and

$latex h(x) = v$

** Step 3:** Get the derivative of the outer function $latex g(v)$, which must use the derivative of the tangent squared function, in terms of $latex v$.

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

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

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

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

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

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

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

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

Given the function

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

its graph shows

And as we know by now, by deriving $latex f(x) = \tan^{2}{(x)}$, we get

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

which if graphed, shows

Illustrating both graphs in one, we have

Looking at the differences between these functions based on those graphs, you can see that the original function $latex f(x) = \tan^{2}{(x)}$ 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 exists within the range of

$latex [0,\infty)$

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 exists within the range of

$latex (-\infty,\infty)$

## Examples

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

### EXAMPLE 1

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

**Solution:** Analyzing the given tangent squared function, it is only a square of a tangent of a single angle $latex \beta$. Therefore, we can use the first method to derive this problem.

**Step 1:** Analyze if the square of tangent of $latex \beta$ is a function of $latex \beta$. In this problem, it is. Hence, proceed to step 2.

**Step 2: **Directly apply the derivative formula of the tangent 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:** $latex G(x) = \tan^{2}{(7x^2-3)}$

**Solution:** Analyzing the given tangent squared function, it is a square of a tangent of a polynomial function. Therefore, we can use the second method to derive this problem.

**Step 1:** Express the tangent squared function as $latex G(x) = \tan^{2}{(v)}$, where $latex v$ represents any function other than *x*. In this problem,

$latex v = 7x^2-3$

We will substitute this later as we finalize the derivative of the problem.

**Step 2:** Consider $latex \tan{(v)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. For this problem, we have

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

and

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

**Step 3:** Get the derivative of the outer function $latex g(v)$, which must use the derivative of the tangent squared function, in terms of $latex v$.

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

**Step 4:** Get the derivative of the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a polynomial function, we will use power rule and sum/difference of derivatives to derive $latex v$.

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

$latex \frac{d}{dx}(h(x)) = 14x$

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

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

$latex \frac{dy}{dx} = 2\tan{(v)}\sec^{2}{(v)} \cdot 14x$

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

$$\frac{dy}{dx} = 2\tan{(v)}\sec^{2}{(v)} \cdot 14x$$

$$\frac{dy}{dx} = 2\tan{(7x^2-3)}\sec^{2}{(7x^2-3)} \cdot 14x$$

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

$$\frac{dy}{dx} = 2\cdot14x \cdot \tan{(7x^2-3)}\sec^{2}{(7x^2-3)}$$

$$\frac{dy}{dx} = 28x\tan{(7x^2-3)}\sec^{2}{(7x^2-3)}$$

And **the final answer is:**

$$G'(x) = 28x\tan{(7x^2-3)}\sec^{2}{(7x^2-3)}$$

## 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 Secant Squared, sec^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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