91Ó°ÊÓ

Implicit differentiation is a useful technique when dealing with equations where it's hard to isolate one variable. For example, in our given problem, we have the equation \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{6} \), which involves both \( x \) and \( y \) intertwined. Instead of solving for \( y \) explicitly in terms of \( x \), we can differentiate both sides of the equation with respect to \( x \) directly.

• Differentiate each term: Apply differentiation to both \( \sin^{-1} x \) and \( \sin^{-1} y \) treating \( y \) as a function of \( x \).
• Treat \( y \) as a function: While differentiating \( \sin^{-1} y \), remember to apply the chain rule because \( y \) depends on \( x \).

This approach allows us to find \( \frac{dy}{dx} \) even when \( y \) can't be isolated.

The arcsin function, denoted as \( \sin^{-1} \, x \), is the inverse of the sine function restricted to a specific range, \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). It is important in this context as it directly relates to both \( x \) and \( y \) in our equation.

• Understanding Bounds: Since \( \sin^{-1} x \) returns values between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), we know \( x \) and \( y \) must also fall within this range for real number outputs.
• Inverse Nature: Remember, arcsin returns the angle whose sine is the given value, which makes it crucial in trigonometric equations like ours.

The behavior and derivative of the arcsin function play key roles in calculations involving curves defined by trigonometric relationships.

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. In our situation, when differentiating \( \sin^{-1} y \) with respect to \( x \), the chain rule helps particularly because \( y \) is implicitly a function of \( x \).
The basics of chain rule involve:

• Differentiate the outer function: First, take the derivative of \( \sin^{-1} y \) with respect to \( y \), which is \( \frac{1}{\sqrt{1-y^2}} \).
• Multiply by the derivative of the inner function: Multiply this result by \( \frac{dy}{dx} \), because \( y \) itself is a function of \( x \).

Thus, combining these results through the chain rule helps us find \( \frac{dy}{dx} \). This rule is indispensable when working with implicit differentiation.

Solving calculus problems often requires a structured approach, like finding the equation of a tangent line. The solution involves several interrelated steps. Here’s how you can tackle such problems:
1. **Understand the Problem**: Clearly state what you need to find. In this exercise, it was the tangent line's equation at a point.
2. **Set Up for Differentiation**: Identify the differentiation method needed—in our case, implicit differentiation.
3. **Differentiate**: Carefully apply the appropriate rules, such as the chain rule, to differentiate.
4. **Solve and Simplify**: Work algebraically to solve for the derivative and simplify where possible to isolate \( \frac{dy}{dx} \).
5. **Use Point-Slope Formula**: With the derivative and point, use the tangent line's formula \( y - y_1 = m(x - x_1) \) to find the equation.

• By practicing these steps regularly, you develop a problem-solving mindset that will help tackle more complex calculus challenges.
• Always double-check your work for mistakes, especially in differentiation and algebraic manipulation.