The derivative of a cotangent function of double angles results in a composite function. **The derivative of the cotangent of 2x is equal to minus two cosecant of 2x, -2csc(2x).** This derivative can be found using the chain rule and the derivative of cotangent.

In this article, we’ll look at how to find the composite function cotangent of double angles. We will go over some fundamentals, definitions, formulas, graph comparisons of underived and derived *cot(2x)*, proofs, derivation techniques, and some examples.

## Proof of the derivative of the cotangent of double angles using chain rule

In the composite function *cot(2x)*, the trigonometric function cotangent will be the outer function *f(u)*, while the monomial *2x* will be the inner function *g(x)*.

Because the cotangent of a double angle is a composite function, the chain rule formula is utilized as the basis to derive it.

As a prerequisite for this topic, please review the chain rule formula by looking at this article: Chain Rule of derivatives. You may also check out this article for the proof of the derivative of the cotangent function: Derivative of Cotangent, cot(x).

Let’s have the derivative of the function

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

We can determine the two functions that comprise *F(x)*. In this instance, there is a trigonometric function cotangent and a monomial. The above function shows that cotangent function is clearly the outer function, whereas the monomial *2x* is the inner function. We may configure the outer function as follows:

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

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 cotangent in terms of *u*, we have

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

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

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

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

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

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

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

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

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

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

In this scenario, we choose not to use the double angle trigonometric identity for cotangent since it complicates the derivative formula. As a result, we arrive at the *cot(2x)* derivative formula.

$latex \frac{d}{dx} \cot{(2x)} = -2\csc^{2}{(2x)}$

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

As previously stated, the cotangent of a double angle is a combination of the trigonometric function cotangent and the monomial 2x. This function is straightforward to calculate, and instead of using the chain rule method all the time, we may just apply the proven derivative formula for the cotangent of a double angle.

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

$latex \frac{d}{dx} \left( \cot{(2x)} \right) = -2\csc^{2}{(2x)}$ |

**Step 1:** Analyze if $latex \cot{(2x)}$ is a function of the same variable $latex x$ or *f(x)*. If $latex \cot{(2x)}$ 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 cotangent of a double angle.

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

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

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

$latex \frac{d}{dx} \left( \cot{(2v)} \right) = -2\csc^{2}{(2v)} \cdot \frac{d}{dx} (v)$ |

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

*x*.

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

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

and

$latex h(x) = v$

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

$latex \frac{d}{du} \left( \cot{(2v)} \right) = -2\csc^{2}{(2v)}$

** 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 $latex h(x)$

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

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

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

Having the function

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

its graph illustrates

And as we know by now, deriving $latex f(x) = \cot{(2x)}$ will lead to

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

which if graphically illustrated is

Illustrating both graphs in one, we have

Based on the differences between these graphs, you can conclude that the original function $latex f(x) = cot{(2x)}$ has a domain of

$$\left(-\frac{3\pi}{2},-\pi\right) \cup \left(-\pi,-\frac{\pi}{2}\right) \cup \left(\frac{-\pi}{2},0\right) \cup \left(0,\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\pi\right) \cup \left(\pi,\frac{3\pi}{2}\right)$$

*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*

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

$$\left(-\frac{3\pi}{2},-\pi\right) \cup \left(-\pi,-\frac{\pi}{2}\right) \cup \left(\frac{-\pi}{2},0\right) \cup \left(0,\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\pi\right) \cup \left(\pi,\frac{3\pi}{2}\right)$$

*within the finite intervals of*

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

and exists within the range of

$latex (-\infty,-2]$

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

The graphs below demonstrate the difference between $latex \cot{(2x)}$ and $latex \cot{(x)}$.

and in terms of their derivatives

## Examples

The following are some examples of calculating the cotangent of a double angle using either the first or second method, depending on which is most appropriate.

### EXAMPLE 1

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

**Solution:**

**Step 1:** Analyzing the given cotangent 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 cotangent of a double angle in terms of $latex \beta$. Since no further simplification is needed, **the final answer is:**

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

### EXAMPLE 2

**Derive:** $latex G(x) = \cot{(2e^x})$

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

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

$latex v = e^x$

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

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

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

and

$latex h(x) = v = e^x$

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

$latex \frac{d}{du} \left( \cot{(2v)} \right) = -2\csc^{2}{(2v)}$

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

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

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

**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 $latex h(x)$

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

$latex \frac{dy}{dx} = (-2\csc^{2}{(2v)}) \cdot (e^x)$

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

$latex \frac{dy}{dx} = (-2\csc^{2}{(2v)}) \cdot (e^x)$

$latex \frac{dy}{dx} = -(2\csc^{2}{(2(e^x))}) \cdot (e^x)$

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

$latex G'(x) = -2e^x\csc^{2}{(2e^x)}$

## See also

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

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