Derivative of Inverse Trigonometric Functions arccsc(x) using Implicit Differentiation

In this tutorial, we will gain a deep understanding on solving problems related to differential calculus. This comprehensive guide offers a structured approach on effectively solving mathematical problems related to derivatives of an inverse trigonometric functions using the method of implicit differentiation. Implicit differentiation is an effective technique to solve for an inverse trigonometric function as we take the derivative of both sides with respect to x.   Let's immerse ourselves in the world of numbers and dominate the wonders of mathematics.

$$ arccsc (x)$$

Let us complete the problem by equating arccsc(x) to y in order to calculate and manipulate the problem as  y is dependent to the value of x and we can easily graph the equation:

$$ y = arccsc(x)$$

Isolate x by taking the cosecant of both sides:

$$ {\color{#FF0000} csc}  y =  {\color{#FF0000} csc} (arccsc(x))$$

Cancel arc cosecant with cosecant as the two trigonometric functions are opposite, resulting to x:

$$csc (y) = \bcancel{csc} (\bcancel{arccsc}(x)) $$

 $$ csc (y) = x$$

Using implicit differentiation, find the derivative of both csc y using chain rule and x using power rule with respect to x:

$$\color{#0000FF} - csc(y)\medspace cot(y) \color{#000000} \cdot \color{#008000} \dfrac{dy}{dx}   \color{#000000} = 1$$

Isolate terms that has dy/dx:

$$\dfrac{\bcancel{-csc(y)cot(y)}\dfrac{dy}{dx}}{\color{#FF0000} \bcancel{-csc(y)cot(y)}}  = - \dfrac{1}{\color{#FF0000} csc(y)\medspace cot(y)}  $$

$$ \dfrac{dy}{dx} = -\dfrac{1}{csc(y)\medspace cot(y))} $$

Find the values of csc(y) and cot(y) using Pythagorean theorem:

$$ csc\medspace \theta = \dfrac{\color{#FF0000} hyp}{\color{#0000FF} opp}\enspace\enspace \enspace\longrightarrow\enspace\enspace  csc\medspace (y) = \dfrac{\color{#FF0000} x}{\color{#0000FF} 1} $$

$$ cot\medspace \theta = \dfrac{\color{#008000} adj}{\color{#0000FF} opp}\enspace\enspace \enspace\longrightarrow\enspace\enspace  cot\medspace (y) = ? $$

Let's us solve and complete the sides of the right triangle in order for us to solve cot(y):

$$a^2 + b^2 = c^2 $$

$$(1)^2 + b^2  = (x)^2 $$

$$1 + b^2  = x^2 $$

$$b^2  = x^2 - 1 $$

$$\sqrt{b^2} = \sqrt{x^2 - 1}$$

$$b = \sqrt{x^2 - 1}$$

Distribute the values of the adjacent and opposite sides of the triangle to calculate for the value of cot(y):

$$ cot\medspace (y)= \dfrac{\color{#008000} \sqrt{x^2 - 1}}{\color{#0000FF} 1} $$

$$ cot\medspace (y)= \sqrt{x^2 - 1}$$

Distribute the computed values of csc(y) and cot(y) from the previous equation:

$$ \dfrac{dy}{dx} = -\dfrac{1}{csc(y)\medspace cot(y))} \enspace\enspace \enspace\longrightarrow\enspace\enspace   \dfrac{dy}{dx} = -\dfrac{1}{x\cdot\sqrt{x^2 - 1}}$$

The derivative of arccsc(x) is:

$$\dfrac{d}{dx}(arccsc(x)) = -\dfrac{1}{x\medspace\sqrt{x^2 - 1}}$$   

Practicing and understanding how to solve derivatives of an inverse trigonometric functions will help us further improve our mathematical skills. It is a valuable skill for anyone studying calculus or pursuing fields that involves advanced mathematics. 

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