The **Derivative of Cosecant** is one of the first transcendental functions introduced in Differential Calculus (*or Calculus I*). **The derivative of the cosecant function is equal to minus cosecant times cotangent, -csc(x) cot(x)**. We can prove this derivative using limits and trigonometric identities.

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

## Proof of the Derivative of the Cosecant Function

The trigonometric function *cosecant* of an angle is defined as the ratio of hypothenuse 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 cosecant of angle *A* can be evaluated as

$latex \csc{(A)} = \frac{c}{a}$

where *A* is the angle, *c* is the hypothenuse, and *a* is its opposite side.

Before learning the proof of the derivative of the cosecant 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

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

we have

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

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

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

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

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

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

Looking at the re-arranged numerator, we can try to apply the *product-sum identities of sine*.

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ 2\cos{\left(\frac{x+(x+h)}{2}\right)} \sin{\left(\frac{x-(x+h)}{2}\right)} }{h\sin{(x+h)}\sin{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ 2\cos{\left(\frac{x+x+h}{2}\right)} \sin{\left(\frac{x-x-h}{2}\right)} }{h\sin{(x+h)}\sin{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ 2\cos{\left(\frac{2x+h}{2}\right)} \sin{\left(\frac{-h}{2}\right)} }{h\sin{(x+h)}\sin{(x)}} \right)}$$

Based on the trigonometric identities of a sine of a negative angle, it is equal to negative sine of the positive form of the same angle. Applying this to the second multiplicand of the numerator, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ 2\cos{\left(\frac{2x+h}{2}\right)} \cdot \left( -\sin{\left(\frac{h}{2}\right)} \right) }{h\sin{(x+h)}\sin{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -2\cos{\left(\frac{2x+h}{2}\right)} \sin{\left(\frac{h}{2}\right)} }{h\sin{(x+h)}\sin{(x)}} \right)}$$

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

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ -\cos{\left(\frac{2x+h}{2}\right)} \cdot 2\sin{\left(\frac{h}{2}\right)} }{\sin{(x+h)}\sin{(x)} \cdot h} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+h}{2}\right)} }{ \sin{(x+h)}\sin{(x)} } \cdot \frac{ 2\sin{\left(\frac{h}{2}\right)} }{h} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+h}{2}\right)} }{ \sin{(x+h)}\sin{(x)} } \right)} \cdot \lim \limits_{h \to 0} {\left( \frac{ 2\sin{\left(\frac{h}{2}\right)} }{h} \right)}$$

Applying some rules of fraction to the second multiplicand, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+h}{2}\right)} }{ \sin{(x+h)}\sin{(x)} } \right)} \cdot \lim \limits_{h \to 0} {\left( \frac{ \sin{\left(\frac{h}{2}\right)} }{ \frac{h}{2} } \right)}$$

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 $latex \sin{\left(\frac{h}{2}\right)}$ over $latex \frac{h}{2}$. Applying, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+h}{2}\right)} }{ \sin{(x+h)}\sin{(x)} } \right)} \cdot \lim \limits_{h \to 0} {1}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+h}{2}\right)} }{ \sin{(x+h)}\sin{(x)} } \right)}$$

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

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x+(0)}{2}\right)} }{ \sin{(x+(0))}\sin{(x)} } \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{-\cos{\left(\frac{2x}{2}\right)} }{ \sin{(x)}\sin{(x)} } \right)}$$

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

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

Applying some trigonometric identities to simplify the derivative formula by the use of *defining relation identities*, we have

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

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

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

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

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

## How to derive a Cosecant Function?

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

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

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

**Step 1:** Analyze if the cosecant of an angle is a function of that same angle. For example, if the right-hand side of the equation is $latex \csc{(x)}$, 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 $latex \csc{(x)}$ 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 cosecant function

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

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

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

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

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

*x*.

** Step 2:** Consider $latex \csc{(u)}$ 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

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

and

$latex g(x) = u$

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

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

** 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 $latex g(x)$

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

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

** 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 Cosecant *x* VS. The Derivative of Cosecant *x*

Given the function

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

the graph is illustrated as

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

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

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

$latex (-\infty,-1] \cup [1,\infty)$

whereas the derivative $latex f'(x) = -\csc{(x)}\cot{(x)}$ 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*

## Examples

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

### EXAMPLE 1

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

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

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

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

### EXAMPLE 2

**Derive:** $latex F(x) = \csc{\left(5-10x^2 \right)}$

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

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

$latex u = 5-10x^2$

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

**Step 2:** Consider $latex \csc{(u)}$ 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

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

and

$latex g(x) = u = 5-10x^2$

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

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

**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 $latex u$.

$$\frac{d}{dx}(g(x)) = \frac{d}{dx} \left(5-10x^2 \right)$$

$$\frac{d}{dx}(g(x)) = -20x$$

**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 $latex g(x)$

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

$$\frac{dy}{dx} = -\csc{(u)}\cot{(u)} \cdot (-20x)$$

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

$$\frac{dy}{dx} = -\csc{(u)}\cot{(u)} \cdot (-20x)$$

$$\frac{dy}{dx} = -\csc{(5-10x^2)}\cot{(5-10x^2)} \cdot (-20x)$$

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

$$\frac{dy}{dx} = \csc{(5-10x^2)}\cot{(5-10x^2)} \cdot 20x$$

$$\frac{dy}{dx} = 20x \csc{(5-10x^2)}\cot{(5-10x^2)}$$

And **the final answer is:**

$$F'(x) = 20x \csc{(5-10x^2)}\cot{(5-10x^2)}$$

or

$$F'(x) = 20x \csc{(5(1-2x^2))}\cot{(5(1-2x^2))}$$

## See also

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