Problem 328 - Calculus Volume 1

Question

For the following exercises, use implicit differentiation to determine \(y^{\prime} .\) Does the answer agree with the formulas we have previously determined? $$ x=\sin y $$

Step-by-Step Solution

Answer

Yes, the answer \(y' = \frac{1}{\sqrt{1-x^2}}\) agrees with the formulas.

Step 1: Differentiate Both Sides with Respect to x

Start by differentiating both sides of the equation \(x = \sin y\) with respect to \(x\). For the left side, the derivative is \(1\). For the right side, use the chain rule, remembering that the derivative of \(\sin y\) is \(\cos y\cdot \frac{dy}{dx}\) or \(\cos y \cdot y'\).

Step 2: Solve for y'

Now, set up the equation from the derivatives: \(1 = \cos y \cdot y'\). We now isolate \(y'\) by dividing both sides by \(\cos y\): \(y' = \frac{1}{\cos y}\).

Step 3: Simplify the Derivative Expression

Since \(\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}\) (because \(x = \sin y\)), substitute \(\cos y\) in the expression for \(y'\): \[ y' = \frac{1}{\sqrt{1-x^2}} \].

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate composite functions. When a function is composed of multiple layers, like an outer function and an inner function, we use the Chain Rule to break them down into manageable parts for differentiation.
In this exercise, we had the function \( x = \sin y \), where \( y \) is implicitly a function of \( x \). We need to differentiate with respect to \( x \), even though \( y \) isn't explicitly expressed as a function of \( x \).
Using the Chain Rule, the derivative of \( \sin y \) with respect to \( x \) becomes \( \cos y \cdot \frac{dy}{dx} \).

• First, differentiate the outer function: the derivative of \( \sin y \) with respect to \( y \) is \( \cos y \).
• Next, multiply by the derivative of the inner function \( y \) with respect to \( x \), which we denote as \( \frac{dy}{dx} \) or \( y' \).
• This produces the result \( \cos y \cdot y' \), perfectly illustrating the application of the Chain Rule in implicit differentiation.

Trigonometric Functions

Trigonometric functions relate the angles of a triangle to the lengths of its sides. They are periodic and have derivatives that are closely connected, which is essential in calculus.
In this problem, we start with \( x = \sin y \). The sine function, \( \sin \), is periodic and has a well-known derivative. The derivative of \( \sin y \) with respect to \( y \) is \( \cos y \). Therefore, as part of differentiating \( x = \sin y \), we needed to use \( \cos y \).
The connection between sine and cosine derives from how triangles behave and is fundamental to solving such problems. Here are the crucial aspects of trigonometric derivatives used:

• \( \sin y \) triggers a move to its derivative, \( \cos y \).
• This derivative is central to converting trigonometric expressions in hybrid equations like \( x = \sin y \).

This close relationship makes trigonometric functions indispensable when dealing with implicit differentiation, particularly within composite functions.

Derivatives

Derivatives are central to calculus, representing rates of change. In this problem, determining \( y' \) involves performing implicit differentiation to find the derivative of \( y \) with respect to \( x \).
Start by finding the derivative of each side of the given equation \( x = \sin y \).

• Differentiating the left side, \( x \), with respect to \( x \) gives \( 1 \).
• Differentiating the right side, \( \sin y \), while accounting for our earlier discussion using the Chain Rule, yields \( \cos y \cdot \frac{dy}{dx} \).

This establishes the relationship \( 1 = \cos y \cdot y' \).
To isolate \( y' \), divide both sides by \( \cos y \) to solve for \( y' \). This results in \( y' = \frac{1}{\cos y} \).
Finally, by recognizing \( \cos y = \sqrt{1 - x^2} \) from the trigonometric identity because \( x = \sin y \), we simplify this to \( y' = \frac{1}{\sqrt{1 - x^2}} \). This expression shows how derivatives capture complex relationships in functions, especially using implicit forms.