Derivative of Cosecant, csc(x) – Formula, Proof, and Graphs

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.

CALCULUS
Derivative of cosecant csc(x)

Relevant for

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

See proof

CALCULUS
Derivative of cosecant csc(x)

Relevant for

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

See proof

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

Right-Triangle-ABC-abc-min

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

graph-of-fx-cscx-min

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

graph of derivative of csc(x)-min

Illustrating both graphs in one, we have

graph of csc(x) and its derivative-min

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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