Derivative of Cotangent, cot(x) – Formula, Proof, and Graphs

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, -csc2(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.

See proof

CALCULUS

Relevant for

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

See proof

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.

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.

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.

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

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

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

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

or

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