91Ó°ÊÓ

Implicit differentiation is a technique used when a function is not explicitly defined in terms of an independent variable. In our exercise, instead of having a direct relationship like \( y = f(x) \), we work with an equation involving both \( x \) and \( y \). Here, the relationship is given implicitly as \( x = \sin y \).
To differentiate implicitly, you treat both \( x \) and \( y \) as variables dependent on time or another parameter, and differentiate each term with respect to \( x \). By doing this, one can find derivatives even when the function forms are not neatly organized.

• Differentiate both sides of the equation \( x = \sin y \) with respect to \( x \). This involves recognizing that \( y \) is not a constant and its derivative, \( \frac{dy}{dx} \), needs to be considered.
• In the expression \( \sin y \), apply implicit differentiation, treating \( y \) as a function of \( x \). The result is: \( 1 = \cos y \cdot \frac{dy}{dx} \).

Implicit differentiation can be a crucial tool for solving problems involving inverse functions, especially when these are not easily expressible in a simple form.

The chain rule is fundamental in calculus and is particularly useful when dealing with composite functions, or functions within functions. In this problem, it's used because we must differentiate a trigonometric function of \( y \), which itself depends on \( x \).
When applying the chain rule, if you have a composite function \( z = f(g(x)) \), the derivative is given by:

• \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \)

In our case, \( f(y) = \sin y \), and thus we have to apply the chain rule to differentiate \( \sin y \) with respect to \( x \), knowing that \( y \) is a function of \( x \).

• The derivative of \( \sin y \) with respect to \( y \) is \( \cos y \).
• Multiply by the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} \).

So, the application of the chain rule yields \( \cos y \cdot \frac{dy}{dx} \), which is an important step in solving our problem.

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. They are often used to simplify expressions or calculations in trigonometry.
In our exercise, the key identity used is derived from the Pythagorean identity. Knowing \( x = \sin y \), we utilize the identity:

• \( \cos^2 y + \sin^2 y = 1 \)

From this, we can solve for \( \cos y \):

• \( \cos y = \sqrt{1 - \sin^2 y} \)
• Substituting \( \sin y = x \), we get \( \cos y = \sqrt{1 - x^2} \).

This identity allows us to express \( \cos y \) in terms of \( x \), making it possible to replace \( \cos y \) in the derivative obtained through implicit differentiation.
Employing this identity bridges the gap between the trigonometric and algebraic expressions and is instrumental in ultimately showing that \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \).