The derivative of a tangent function of double angles results in a composite function. **The derivative of the tangent of 2x is equal to two times secant squared of 2x, 2sec ^{2}(2x).** We can find this derivative using the chain rule and the derivative of tangent.

In this post, we’ll look at how to find the composite function tangent of double angles. We will cover some principles, definition, formula, graph comparison of underived and derived *tan(2x)*, proof, derivation procedures, and a few examples.

## Proof of The Derivative of The Tangent of Double Angles Using Chain Rule

Because it is a composite function, the chain rule formula is used as the foundation to derive the tangent of a double angle. The trigonometric function tangent will be the outer function *f(u)* in the composite function *tan(2x)*, whilst the monomial *2x* will be the inner function *g(x)*.

As a precondition for this topic, you should review the chain rule formula by visiting this article: Chain Rule of derivatives. In addition, you can visit this article for the proof of tangent function derivative: Derivative of Tangent, tan(x).

Suppose we are asked to get the derivative of

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

We can identify the two functions that make up *F(x)*. There is a trigonometric function tangent and a monomial in this scenario. It is evident that the given tangent function is the outer function, while the monomial *2x* is the inner function. We can set the outer function as

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

where

$latex u = 2x$

Setting the monomial *2x* as the inner function of *f(u)* by denoting it as *g(x)*, we have

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

$latex g(x) = 2x$

$latex u = g(x)$

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

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

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

Deriving the inner function *g(x)* using power rule since it is a monomial, we have

$latex g(x) = 2x$

$latex g'(x) = 2$

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

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

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

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

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

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

In this case, we prefer not to apply the double angle trigonometric identity for tangent as it will make the derivative formula less simplified. Therefore, this gets us to the *tan(2x)* derivative formula

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

## How to derive the Tangent of a Double Angle?

As mentioned above, the tangent of a double angle is a composite function of the trigonometric function tangent and the monomial *2x*. This function is simple to derive, and instead of utilizing the chain rule method all the time, we may just use the proven derivative formula for the tangent of a double angle.

### METHOD 1: When the tangent of a double angle *2x* is to be derived in terms of the same variable *x*.

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

**Step 1:** Analyze if $latex \tan{(2x)}$ is a function of the same variable $latex x$ or *f(x)*. If $latex \tan{(2x)}$ is a function of other variables such as *f(t) *or *f(y)*, it will use implicit differentiation which is out of the scope of this article.

** Step 2:** Directly apply the proven derivative formula of the tangent of a double angle.

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

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

### METHOD 2: When the given is a tangent of a function $latex v \times 2$ and to be derived in terms of *x.*

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

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

*x*.

** Step 2:** Consider $latex \tan{(2x)}$ 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

$latex g(v) = \tan{(2v)}$

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 of a double angle, in terms of $latex v$.

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

** 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 $latex h(x)$

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

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

** 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 *tan(2x)* VS. its derivative

Given the function

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

its graph is shown as

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

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

which if graphed, is shown as

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 $latex f(x) = \tan{(2x)}$ has a domain o

$latex \left(-\frac{3\pi}{4},-\frac{\pi}{4}\right) \cup \left(-\frac{\pi}{4},\frac{\pi}{4}\right) \cup \left(\frac{\pi}{4},\frac{3\pi}{4}\right)$

*within the finite intervals of*

$latex \left(-\frac{3\pi}{4},\frac{3\pi}{4}\right)$

and exists within the range of

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

whereas the derivative $latex f'(x) = 2\sec^{2}{(2x)}$ has a domain of

$latex \left(-\frac{3\pi}{4},-\frac{\pi}{4}\right) \cup \left(-\frac{\pi}{4},\frac{\pi}{4}\right) \cup \left(\frac{\pi}{4},\frac{3\pi}{4}\right)$

*within the finite intervals of*

$latex \left(-\frac{3\pi}{4},\frac{3\pi}{4}\right)$

and exists within the range of

$latex [2,\infty)$

### Graph Comparison between *tan(2x)* and *tan(x)* as well as their derivatives

The graphs shown below illustrates the difference between $latex \tan{(2x)}$ and $latex \tan{(x)}$

and in terms of their derivatives

## Examples

The following are some examples of deriving a tangent of a double angle by using either the first or the second method, whichever is more applicable.

### EXAMPLE 1

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

**Solution:**

**Step 1:** Analyzing the given tangent of a double angle, it is to be derived in terms of $latex \beta$. Therefore, we can use the first method to derive this problem.

**Step 2: **Directly apply the derivative formula of the tangent of a double angle in terms of $latex \beta$. Since no further simplification is needed, **the final answer is:**

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

### EXAMPLE 2

**Derive:** $latex G(x) = \tan{(2\ln{(x)})}$

**Solution:** Analyzing the given tangent of a function *times* two, it is a tangent of a logarithmic function multiplied by two. Therefore, we can use the second method to derive this problem.

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

$latex v = \ln{(x)}$

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

**Step 2:** Consider $latex \tan{(2v)}$ 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

$latex g(v) = \tan{(2v)}$

and

$latex h(x) = v = \ln{(x)}$

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

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

**Step 4:** Get the derivative of the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a logarithmic function, we will use the derivative of logarithmic functions to derive $latex v$.

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

$latex \frac{d}{dx}(h(x)) = \frac{1}{x}$

**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 $latex h(x)$

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

$latex \frac{dy}{dx} = 2\sec^{2}{(2v)} \cdot \frac{1}{x}$

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

$latex \frac{dy}{dx} = 2\sec^{2}{(2v)} \cdot \frac{1}{x}$

$latex \frac{dy}{dx} = 2\sec^{2}{(2(\ln{(x)}))} \cdot \frac{1}{x}$

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

$latex G'(x) = \frac{2\sec^{2}{(2\ln{(x)})}}{x}$

## See also

Interested in learning more about the derivatives of trigonometric functions of 2x? Take a look at these pages:

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