The Derivative of Cotangent is one of the first transcendental functions introduced in Differential Calculus (*or Calculus I*). **The derivative of the cotangent function is equal to minus cosecant squared, -csc ^{2}(x).** This derivative can be proved using limits and trigonometric identities.

In this article, we will discuss how to derive the trigonometric function cotangent. We will cover brief fundamentals, its formula, a graph comparison of cotangent and its derivative, a proof, methods to derive, and a few examples.

##### CALCULUS

**Relevant for**…

Learning about the proof and graphs of the derivative of cotangent.

##### CALCULUS

**Relevant for**…

Learning about the proof and graphs of the derivative of cotangent.

## Proof of the Derivative of the Cotangent Function

The trigonometric function *cotangent* of an angle is defined as the ratio of adjacent side to the opposite 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 cotangent of angle *A* can be evaluated as

$latex \cot{(A)} = \frac{b}{a}$

where *A* is the angle, *b* is its adjacent side, and *a* is its opposite side.

Before learning the proof of the derivative of the cotangent 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

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

we have

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

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

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

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

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

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

Looking at the re-arranged numerator, we can try to apply the *sum and difference identities for sine and cosine*, also called *Ptolemy’s identities*.

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

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

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

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 our numerator, we have

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

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

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

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

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

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

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

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

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

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

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

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

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

We know that by the *defining relation identities*, the reciprocal of the trigonometric function sine is cosecant. Applying, we have

$$\frac{d}{dx} f(x) = – \left( \frac{1}{\sin{(x)}} \cdot \frac{1}{\sin{(x)}} \right)$$

$$\frac{d}{dx} f(x) = – (\csc{(x)} \cdot \csc{(x)})$$

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

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

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

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

## How to derive a Cotangent Function?

The derivative process of a cotangent function is very straightforward assuming you have already learned the concepts behind the usage of the cotangent function and how we arrived to its derivative formula.

### METHOD 1: Derivative of Cotangent of any angle *x* in terms of the same angle *x*.

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

**Step 1:** Analyze if the cotangent of an angle is a function of that same angle. For example, if the right-hand side of the equation is $latex \cot{(x)}$, then check if it is a function of the same angle *x* or *f(x)*. After this, proceed to Step 2 until you complete the derivation steps.

* Note:* If $latex \cot{(x)}$ 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 derivative formula of the cotangent function

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

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

### METHOD 2: Derivative of Cotangent of any function *u* in terms of *x.*

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

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

*x*.

** Step 2:** Consider $latex \cot{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. Hence we have

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

and

$latex g(x) = u$

** Step 3:** Get the derivative of the outer function $latex f(u)$, which must use the derivative of the cotangent function, in terms of $latex u$.

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

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

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

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

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

** Step 6:** Substitute $latex u$ into $latex f'(u)$

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

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

Given the function

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

the graph is illustrated as

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

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

which is illustrated graphically as

Illustrating both graphs in one, we have

Analyzing the differences of these functions through these graphs, you can observe that the original function $latex f(x) = \cot{(x)}$ has a domain of

$$(-2\pi,-\pi) \cup (-\pi,0) \cup (0,\pi) \cup (\pi,2\pi)$$

*within the finite intervals of*

$latex (-2\pi,2\pi)$

and exists within the range of

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

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

$$(-2\pi,-\pi) \cup (-\pi,0) \cup (0,\pi) \cup (\pi,2\pi)$$

*within the finite intervals of*

$latex (-2\pi,2\pi)$

and exists within the range of

$latex (-\infty,-1]$ or $latex y \leq -1$

## Examples

Below are some examples of using either the first or second method in deriving a cotangent function.

### EXAMPLE 1

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

**Solution:** We can see that this is only a cotangent of a single angle $latex \beta$. Therefore, we can use the first method to derive this problem.

**Step 1:** Analyze if the cotangent 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 cotangent function and derive in terms of $latex \beta$. Since no further simplification is needed, **the final answer is:**

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

### EXAMPLE 2

**Derive:** $latex F(x) = \cot{\left(3-6x^2 \right)}$

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

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

$latex u = 3-6x^2$

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

**Step 2:** Consider $latex \cot{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. For this problem, we have

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

and

$latex g(x) = u = 3-6x^2$

**Step 3:** Get the derivative of the outer function $latex f(u)$, which must use the derivative of the cotangent function, in terms of $latex u$.

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

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

$latex \frac{d}{dx}(g(x)) = \frac{d}{dx} \left(3-6x^2 \right)$

$latex \frac{d}{dx}(g(x)) = -12x$

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

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

$latex \frac{dy}{dx} = (-\csc^{2}{(u)}) \cdot (-12x)$

**Step 6:** Substitute $latex u$ into $latex f'(u)$

$latex \frac{dy}{dx} = (-\csc^{2}{(u)}) \cdot (-12x)$

$latex \frac{dy}{dx} = (-\csc^{2}{(3-6x^2)}) \cdot (-12x)$

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

$latex \frac{dy}{dx} = (-\csc^{2}{(3-6x^2)}) \cdot (-12x)$

$latex \frac{dy}{dx} = \csc^{2}{(3-6x^2)} \cdot 12x$

$latex \frac{dy}{dx} = 12x\csc^{2}{(3-6x^2)}$

And **the final answer is:**

$latex F'(x) = 12x\csc^{2}{(3-6x^2)}$

or

$latex F'(x) = 12x\csc^{2}{(3(1-2x^2))}$

## See also

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