91Ó°ÊÓ

Inverse trigonometric functions are the functions that reverse the action of the corresponding trigonometric functions. The primary functions include arcsine, arccosine, and arctangent, which are denoted as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \) respectively.
These functions are used to find the angle that corresponds to a given sine, cosine, or tangent value.
In our problem, we deal with \( \tan^{-1} x \) and \( \cot^{-1} x \). Understanding the geometry behind inverse trigonometric functions is key.
For \( \tan^{-1} x \), if \( \theta = \tan^{-1} x \), then \( \tan \theta = x \), which points us to a right triangle with \( x \) as opposite and \( 1 \) as adjacent.
This leads us to calculate \( \sin(\tan^{-1}x) \) using a triangle's property.

• \( \sin(\theta) = \frac{x}{\sqrt{x^2+1}} \)
• \( \cos(\theta) = \frac{1}{\sqrt{x^2+1}} \)

Similarly, for \( \cot^{-1} x \), if \( \alpha = \cot^{-1} x \), then \( \cot \alpha = x \), resulting in a different right triangle with \( x \) as adjacent.
This helps derive \( \sin(\cot^{-1}x) \), important for the function studdied in this exercise.

Function simplification is an essential step in easing equation solving and making calculations more manageable. It often involves reducing a complex expression to a simpler form without changing its value.
In this exercise, we simplified the function \( f(x) = (\sin(\tan^{-1} x) + \sin(\cot^{-1} x))^2 - 1 \).
Initially, it involves calculating \( \sin(\tan^{-1} x) \) and \( \sin(\cot^{-1} x) \) based on trigonometric principles.

• By substituting \( \sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2+1}} \) and \( \sin(\cot^{-1} x) = \frac{1}{\sqrt{x^2+1}} \) back into the expression, the function simplifies to \( \frac{x+1}{\sqrt{x^2+1}} \).
• Squaring this results in \( \frac{(x+1)^2}{x^2+1} \), and simplifying further leads to \( f(x) = \frac{2x}{x^2+1} \).

This simplification is crucial for derivative calculation and solving the differential equation. A clear understanding of each component simplifies solving complex functions like the one in this exercise.

Derivative calculation involves finding the rate at which a function changes at any given point. It is a key concept in calculus with numerous applications.
In this exercise, the derivative of the function \( y \) with respect to \( x \) was calculated, where \( \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1}(f(x))) \).
The process relies on the chain rule as well as understanding derivatives of inverse functions.For the derivative of \( \sin^{-1}(f(x)) \):

• The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).
• Inserting \( f(x) = \frac{2x}{x^2+1} \), we calculate \( \frac{d}{dx}(f(x)) = \frac{2(1-x^2)}{(x^2+1)^2} \).

Substitute these components back into \( \frac{dy}{dx} \), simplifying to solve the differential.
This process emphasizes the importance of the derivative in connecting function behavior to real-world scenarios, essential in this calculus exercise.