Problem 328 - Calculus Volume 1

Question

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

The derivative is \( y' = \sec y \). Yes, it agrees with the known formulas.

Step 1: Differentiate Both Sides of the Equation with Respect to x

We start with the given equation: \( x = \sin y \). To find \( y' \) (which is \( \frac{dy}{dx} \)), we apply implicit differentiation to both sides of the equation with respect to \( x \). The derivative of \( x \) with respect to \( x \) is 1, and the derivative of \( \sin y \) with respect to \( x \) is \( \cos y \cdot \frac{dy}{dx} \). This gives us:\[ \frac{d}{dx}[x] = \frac{d}{dx}[\sin y] \] Thus, we have:\[ 1 = \cos y \cdot \frac{dy}{dx} \]

Step 2: Solve for y' (dy/dx)

With the equation obtained in Step 1, \( 1 = \cos y \cdot \frac{dy}{dx} \), we solve for \( \frac{dy}{dx} \). Divide both sides by \( \cos y \):\[ \frac{dy}{dx} = \frac{1}{\cos y} \]Since \( \frac{1}{\cos y} \) is the same as \( \sec y \), we can rewrite this as:\[ \frac{dy}{dx} = \sec y \]

Step 3: Verify the Solution Against Known Formula

Using the known trigonometric identity, differentiate \( x = \sin y \) notationally where \( y = \sin^{-1}(x) \). The derivative of \( y = \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \). Notice, however, in this context, \( y = \sin^{-1}(x) \) implies the domain restriction inherently determining \( \cos y = \sqrt{1-x^2} \).Hence the calculation validifies as \( \frac{dy}{dx} = \frac{1}{\cos y} = \sec y \).

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial when working with implicit differentiation, especially when presented with trigonometric equations like \( x = \sin y \). Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable. They can include identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) or \( \sec \theta = \frac{1}{\cos \theta} \). These identities help us simplify and evaluate expressions involving trigonometric functions. In the context of implicit differentiation, these identities become handy. For instance, when we solve for \( y' \) and find \( \frac{dy}{dx} = \sec y \), we rely on the identity \( \sec y = \frac{1}{\cos y} \). This allows us to reframe the derivative in terms of \( \cos y \), which is particularly useful to verify and cross-check our solutions against expected formulas.

Inverse Functions

Inverse functions are functions that reverse the effect of the original function. If \( y = f(x) \), then the inverse function \( x = f^{-1}(y) \) will satisfy the condition \( f(f^{-1}(y)) = y \). In the context of the exercise, \( y = \sin^{-1}(x) \) is the inverse of \( x = \sin y \). Derivatives of inverse trigonometric functions are essential in calculus. For \( y = \sin^{-1}(x) \), the derivative \( \frac{1}{\sqrt{1-x^2}} \) comes from differentiating the inverse function itself. This inverse derivative is pivotal in connecting back to the identity we derived related to secant. Moreover, understanding inverse functions requires us to consider their domain and range. As with any inverse function, when you determine its derivative, like changing \( x = \sin y \) to find \( y' \), it’s important to account for domain restrictions to ensure the function remains valid.

Differentiation Techniques

Differentiation techniques form the foundation of solving calculus problems. This exercise specifically involves one such technique known as implicit differentiation. Unlike explicit differentiation, where functions are solved directly in terms of one variable, implicit differentiation is used when variables are interdependent and not easily separated.To apply implicit differentiation:

• Differentiate both sides of the equation with respect to the independent variable (usually \( x \)).
• Apply the chain rule when differentiating terms involving the dependent variable (typically \( y \)).
• Solve for the derivative \( \frac{dy}{dx} \).

In our exercise, the differentiation of \( \sin y \) with respect to \( x \) results in \( \cos y \cdot \frac{dy}{dx} \) due to the chain rule. This must then be equated with \( 1 \) (the derivative of \( x \) with respect to \( x \)) to solve for \( y' \). This technique is particularly powerful when dealing with complex equations or functions not easily expressed in "\( y =\) form," ensuring that we can still find derivatives and understand the rate of change of one variable with respect to another.