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## Avoid complication being used the indications arccot( x), cot^{-1}( x),

1

/

cot( x)

, as well as cot^{n}( x)

Let’s beginning by clearing up using the various indications in between $latex message {arccot} (x)$, $latex cot ^ {-1} {(x)} $, $latex frac {1} {cot {(x)}} $ as well as $latex cot ^ {n} {(x)} $, because swapping these signs, we can obtain derivation mistakes.

Summing up the meaning of these signs, we have

$ latex message {arccot} (x) = cot ^ {-1} {(x)} $

The signs $latex message {arccot} $ as well as $latex cot ^ {-1} $ are utilized to signify the inverted cotangent. message {arccot} is utilized as the spoken icon of the inverted cotangent feature, while $latex cot ^ {-1} $ is utilized as the mathematical icon of the inverted cotangent feature for a much more official feature.

When it comes to the indication $latex cot ^ {-1} {(x)} $, we need to take into consideration that $latex -1$ is not an algebraic backer of a cotangent. The $latex -1$ utilized for the inverted cotangent stands for that the cotangent is inverted as well as not increased to $latex -1$.

For that reason,

$ latex cot ^ {-1} {(x)} neq frac {1} {cot {(x)}} $

And givens such as $latex cot ^ {2} {(x)} $ or $latex cot ^ {n} {(x)} $, where *n* is any kind of algebraic backer of a non-inverse cotangent, **MUST NOT** make use of the inverted cotangent formula because in these givens, both the 2 as well as any kind of backer *n* are dealt with as algebraic backers of a non-inverse cotangent.

## Evidence of the By-product of the Inverse Cotangent Function

In this evidence, we will primarily make use of the ideas of an appropriate triangular, the Pythagorean theory, the trigonometric features of cotangent as well as cosecant, as well as some fundamental algebra. Much like in the previous number as a recommendation example for a provided right triangular, intend we have that exact same triangular $latex Delta ABC$, however this moment, allow’s transform the variables for a simpler image.

where for every single *one-unit* of a side contrary to *angle y*, there is a side *x* beside *angle y* as well as a hypotenuse equivalent to $latex sqrt {1+ x ^ 2} $*.*

Utilizing these elements of a right-triangle, we can locate the *angle y* by utilizing Cho-Sha-Cao, specifically the cotangent feature by utilizing its nearby as well as contrary sides.

$ latex cot {(theta)} = frac {adj} {opp} $

$ latex cot {(y)} = frac {x} {1} $

$ latex cot {(y)} = x$

Now, we can unconditionally acquire this formula by utilizing the by-product of trigonometric feature of cotangent for the left-hand side as well as power policy for the right-hand side. Doing so, we have

$ latex frac {d} {dx} (cot {(y)}) = frac {d} {dx} (x)$

$ latex frac {d} {dx} (cot {(y)}) = 1$

$ latex frac {dy} {dx} (- csc ^ {2} {(y)}) = 1$

$ latex frac {dy} {dx} = frac {1} {-csc ^ {2} {(y)}} $

$ latex frac {dy} {dx} = -frac {1} {csc ^ {2} {(y)}} $

Getting the cosecant of *angle y* from our provided right-triangle, we have

$ latex csc {(y)} = frac {hyp} {opp} $

$ latex csc {(y)} = frac {sqrt {1+ x ^ 2}} {1} $

Since we require to replace $latex csc {(y)} $ right into $latex csc ^ {2} {(y)} $, we require to make even both sides

$ latex csc ^ {2} {(y)} = left( frac {sqrt {1+ x ^ 2}} {1} right) ^ 2$

$ latex csc ^ {2} {(y)} = left( sqrt {1+ x ^ 2} right) ^ 2$

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

We can after that replace $latex csc ^ {2} {(y)} $ to the implied distinction of $latex cot {(y)} = x$

$ latex frac {dy} {dx} = -frac {1} {csc ^ {2} {(y)}} $

$ latex frac {dy} {dx} = -frac {1} {1+ x ^ 2} $

Therefore, algebraically addressing for the *angle y* as well as obtaining its by-product, we have

$ latex cot {(y)} = x$

$ latex y = frac {x} {cot} $

$ latex y = cot ^ {-1} {(x)} $

$ latex frac {dy} {dx} = frac {d} {dx} left( cot ^ {-1} {(x)} right)$

$ latex frac {dy} {dx} = -frac {1} {1+ x ^ 2} $

which is currently the acquired formula for the inverted cotangent of *x*.

Currently, for the by-product of an inverted cotangent of any kind of feature apart from *x*, we might use the acquired formula of inverted cotangent along with the chain policy formula. By doing so, we have

$ latex frac {dy} {dx} = frac {d} {du} cot ^ {-1} {(u)} cdot frac {d} {dx} (u)$

$ latex frac {dy} {dx} = -frac {1} {1+ u ^ 2} cdot frac {d} {dx} (u)$

where $latex u$ is any kind of feature apart from *x*.

## Chart of Inverse Cotangent *x* VS. The By-product of Inverse Cotangent *x*

Given the feature

$ latex f( x) = cot ^ {-1} {(x)} $

its chart is

And as we understand now, by obtaining $latex f( x) = cot ^ {-1} {(x)} $, we get

$ latex f'( x) = -frac {1} {1+ x ^ 2} $

which is detailed graphically as

Illustrating both charts in one, we have

Analyzing the distinctions of these features with these charts, you can observe that the initial feature $latex f( x) = cot ^ {-1} {(x)} $ has a domain name of

$ latex (- infty, infty)$ or *all genuine numbers*

and exists within the variety of

$ latex (0, pi)$ or $latex 0<< y<< pi$

whereas the acquired $latex f'( x) = -frac {1} {1+ x ^ 2} $ has a domain name of

$ latex (- infty, infty)$ or *all genuine numbers*

and exists within the variety of

$ latex [-1, 0)$ or $latex -1 leq y < 0$

## Examples

Here are some examples of how to derive a composite inverse cotangent function.

### EXAMPLE 1

What is the derivative of $latex f(x) = cot^{-1}(11x)$?

We have a composite inverse cotangent function, so letâ€™s use the chain rule to derive it.

Considering $latex u=11x$ as the inner function, we can write $latex f(u)=tan^{-1}(u)$. Therefore, applying the chain rule, we have:

$$frac{dy}{dx}=frac{dy}{du} frac{du}{dx}$$

$$frac{dy}{dx}=-frac{1}{1+u^2} times 11$$

Substituting $latex u=11x$ back into the function, we have:

$$frac{dy}{dx}=-frac{11}{1+(11x)^2}$$

$$frac{dy}{dx}=-frac{11}{1+121x^2}$$

### EXAMPLE 2

Find the derivative of the function $latex F(x) = cot^{-1}(x^3+5)$

We are going to use the chain rule. Therefore, we write the inverse tangent function as $latex f (u) = cot^{-1}(u)$, where $latex u = x^3+5$.

Now, we calculate the derivative of the outer function $latex f(u)$:

$$frac{d}{du} ( cot^{-1}(u) ) = -frac{1}{1+u^2}$$

Then, we calculate the derivative of the inner function $latex g(x)=u=x^3+5$:

$$frac{d}{dx}(g(x)) = frac{d}{dx}(x^3+5)$$

$$frac{d}{dx}(g(x)) = 3x^2$$

The chain rule tells us that we have to multiply the derivative of the outer function by the derivative of the inner function:

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

$$frac{dy}{dx} = -frac{1}{1+u^2} cdot 3x^2$$

Finally, we substitute $latex u$ back in and simplify:

$$frac{dy}{dx} = -frac{1}{1+(x^3+5)^2} cdot 3x^2$$

$$frac{dy}{dx} = -frac{3x^2}{1+x^6+10x^3+25}$$

$$F'(x) = -frac{3x^2}{x^6+10x^3+26}$$

### EXAMPLE 3

If we have the function $latex f(x) = cot^{-1}(sqrt{x})$, what is its derivative?

The square root function is the inner function. Therefore, since $latex u=sqrt{x}$ is equal to $latex u=x^{frac{1}{2}}$, we have the following derivative:

$$frac{du}{dx}=frac{1}{2}x^{-frac{1}{2}}$$

Since we have $latex f(u)=cot^{-1}(u)$ applying the chain rule gives us:

$$frac{dy}{dx}=frac{dy}{du} frac{du}{dx}$$

$$frac{dy}{dx}=-frac{1}{1+u^2} times frac{1}{2}x^{-frac{1}{2}}$$

Substituting $latex u=sqrt{x}$ and simplifying, we have:

$$frac{dy}{dx}=-frac{1}{1+(sqrt{x})^2} times frac{1}{2}x^{-frac{1}{2}}$$

$$frac{dy}{dx}=-frac{1}{1+x} times frac{1}{2}x^{-frac{1}{2}}$$

$$frac{dy}{dx}=-frac{1}{2sqrt{x}(1+x)}$$

## Practice of derivatives of composite inverse cotangent functions

You have completed the quiz!

## See also

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