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## Proof of The By-product of Cosecant Made Even Feature with making use of Chain Rule

As a requirement, please evaluate the chain policy formula and also its evidence by taking a look at the short article: Chain Rule of derivatives In a similar way, you can evaluate the evidence of the by-product of cosecant feature by seeing this short article: Derivative of Cosecant, csc(x).

Please be advised that

$latex csc ^ {2} {(x)} neq csc {(x ^ 2)} $

To protect against misconception, the very first is a “complete trigonometric feature” increased to the power of 2, whereas the 2nd is a trigonometric feature of “the square of a variable.”

Because it is a composite feature, the chain policy formula is utilized as a much more simple device to verify the by-product of the cosecant made even feature, giving you currently understand exactly how to verify the chain policy formula and also the by-product of a cosecant feature.

Thinking we are asked to discover the acquired of

$latex F(x) = csc ^ {2} {(x)} $

We can recognize both features that comprise F(x). There is a power feature and also a trigonometric feature in this circumstance. To be much more precise, these are a feature increased to a power of 2 and also a trigonometric feature of cosecant, based upon our offered F(x).

For a much better depiction, we can reword it as

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

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

It ends up being noticeable that the enabled feature is the external feature to be taken into consideration, while the cosecant feature, being increased by the enabled feature, is the internal feature. We can set up the external feature as adheres to:

$latex f(u) = u ^ 2$

where

$latex u = csc {(x)} $

The trigonometric cosecant feature, as the internal feature of *f(u)*, will certainly be represented as *g(x)*.

$latex f(u) = f(g(x))$

$latex u = g(x)$

$latex g(x) = csc {(x)} $

Deriving the external feature *f(u)* utilizing the power policy in regards to *u*, we have

$latex f(u) = u ^ 2$

$latex f'(u) = 2u$

Deriving the internal feature *g(x)* utilizing the acquired formula of trigonometric feature *cosecant* in regards to *x*, we have

$latex g(x) = csc {(x)} $

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

Algebraically increasing the by-product of external feature $latex f'(u)$ by the by-product of internal feature $latex g'(x)$, we have

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

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

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

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

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

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

which brings us to the acquired formula of cosecant made even *x*

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

## Relationship in between the by-product of cosecant made even and also cotangent made even, what makes them comparable?

You might ask yourself why

$latex csc ^ {2} {(x)} $

and

$latex cot ^ {2} {(x)} $

have comparable by-products.

According to the Pythagorean formula for cosecants and also cotangents,

$latex csc ^ {2} {(x)} = 1 + cot ^ {2} {(x)} $

If we attempt to acquire both sides of the formula, we have

$latex frac {d} {dx} (csc ^ {2} {(x)}) = frac {d} {dx} (1) + frac {d} {dx} (cot ^ {2} {(x)} )$

Evaluating the by-product of the very first term in the right-hand-side of the formula, where the by-product is no, we have

$latex frac {d} {dx} (csc ^ {2} {(x)}) = 0 + frac {d} {dx} (cot ^ {2} {(x)} )$

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

This is why both the cosecant made even and also the cotangent made even have the very same by-product, due to the Pythagorean formula for cosecants and also cotangents.

## Just how is a Cosecant Settled Feature acquired? *Quicker Methods*

Cosecant made even, as formerly specified, is a composite feature of power and also the trigonometric feature cosecant. Rather than utilizing the chain policy formula over and over again like what we carried out in the evidence, we might just utilize the well established acquired formula for a cosecant made even feature.

### TECHNIQUE 1: By-product of the square of a cosecant of any kind of angle *x* in regards to the very same angle *x*.

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

**Step 1:** Determine whether the cosecant made even of an angle is a feature of the very same angle. As an example, if the right-hand side of the formula is $latex csc ^ {2} {(x)} $, establish if it is a feature of the very same angle *x* or *f(x)*.

* Note:* If $latex csc ^ {2} {(x)} $ is a feature of a various angle or variable, such as

*f(t)*or

*f(y)*, implied distinction will certainly be utilized, which is past the extent of this short article.

** Step 2: ** After that straight use the tested acquired formula of the cosecant made even function

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

If absolutely nothing else can be streamlined, then that is the last service.

### TECHNIQUE 2: By-product of the square of a cosecant of any kind of feature *v* in regards to *x.*

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

** Step 1:** Express the feature as $latex G(x) = csc ^ {2} {(v)} $, where $latex v$ stands for any kind of feature apart from

*x*.

** Action 2:** Deal with $latex csc ^ {2} {(v)} $ as the outdoors feature $latex g(v)$ and also $latex v$ as the internal feature $latex h(x)$ of the composite feature $latex G(x)$. Doing this, we have

$latex g(v) = csc {(v)} $

and

$latex h(x) = v$

** Step 3:** Obtain the external feature $latex g(v)$, and also utilize the by-product of the cosecant made even feature, in regards to $latex v$.

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

** Step 4:** Obtain the internal feature $latex h(x) = v$. Usage whatever suitable by-product policy puts on $latex v$.

** Action 5:** Algebraically increase the by-product of external feature $latex g(v)$ by the by-product of internal feature $latex h(x)$ to entirely use chain rule

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

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

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

** Step 7:** Simplify and also use any kind of feature legislation whenever relevant after that wrap up the solution.

## Chart of Cosecant Made Even *x* VS. The By-product of Cosecant Settled *x*

With the feature

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

it is graphed as

As we currently understand, acquiring $latex f(x) = csc ^ {2} {(x)} $ is

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

which is graphed as

Illustrating both charts in one, we have

By taking a look at the distinctions in between these features basing upon the charts over, you can see that the initial feature $latex f(x) = csc ^ {2} {(x)} $ has a domain name of

$$(-2 pi,-pi) mug (-pi,0) mug (0, pi) mug (pi,2 pi)$$

*within the limited periods of*

$latex (-2 pi,2 pi)$

and exists within the variety of

$latex [1,infty)$

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

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

*within the finite intervals of*

$latex (-2pi,2pi)$

and lies within the range of

$latex (-infty,infty)$

## Examples

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

### EXAMPLE 1

**Derive:** $latex f(beta) = csc^{2}{(beta)}$

**Solution:** After examining the given cosecant squared function, it shows that it is only a square of a cosecant of a single angle $latex beta$. Hence, we can apply the first method to this problem.

**Step 1:** Assess if the square of cosecant $latex beta$ is a function of $latex beta$. It is in this problem. Hence, proceed to step 2.

**Step 2: **Directly apply the derivative formula of the cosecant squared function and derive in terms of $latex beta$. Since no further simplification is needed, **the final answer is:**

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

### EXAMPLE 2

**Derive:** $latex G(x) = csc^{2}{(4-9x^2)}$

**Solution:** After examining the given cosecant squared function, it shows that it is a square of a cosecant of a polynomial function. Hence, we can apply the second method to this problem.

**Step 1:** Express the cosecant squared function as $latex G(x) = csc^{2}{(v)}$, using $latex v$ to represent any function other than *x*, which is the angle of the cosecant squared. In this problem,

$latex v = 4-9x^2$

Let’s substitute this later as we evaluate the derivative of the problem.

**Step 2:** Consider $latex csc{(v)}$ as the outside function $latex g(v)$. Then $latex v$ will be the inner function, denoted as $latex h(x)$ too. For this problem, we have

$latex g(v) = csc{(v)}$

and

$latex h(x) = v = 4-9x^2$

**Step 3:** Derive the outer function $latex g(v)$ using the derivative of the cosecant squared function, in terms of $latex v$.

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

**Step 4:** Derive the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a polynomial function, let’s apply the power rule and sum/difference of derivatives to derive $latex v$.

$latex frac{d}{dx}(h(x)) = frac{d}{dx} left(4-9x^2 right)$

$latex frac{d}{dx}(h(x)) = -18x$

**Step 5:** Evaluate the basic chain rule formula by algebraically multiplying the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$

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

$$frac{dy}{dx} = (-2cot{(v)}csc^{2}{(v)}) cdot (-18x)$$

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

$$ frac{dy}{dx} = (-2cot{(v)}csc^{2}{(v)}) cdot (-18x)$$

$$frac{dy}{dx} = (-2cot{(4-9x^2)}csc^{2}{(4-9x^2)}) cdot (-18x)$$

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

$$frac{dy}{dx} = (-2cot{(4-9x^2)}csc^{2}{(4-9x^2)}) cdot (-18x)$$

$$frac{dy}{dx} = 2cot{(4-9x^2)}csc^{2}{(4-9x^2)} cdot 18x$$

$$frac{dy}{dx} = 36x cot{(4-9x^2)}csc^{2}{(4-9x^2)}$$

And **the final answer is:**

$$G'(x) = 36x cot{(4-9x^2)}csc^{2}{(4-9x^2)}$$

## See also

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