Problem 329 - 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? $$ x=\cos y $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{1}{\sin y}\), matching the known formula.

Step 1: Differentiate Both Sides

Begin by differentiating both sides of the equation with respect to \(x\). This means we take the derivative of \(x = \cos y\) implicitly. The derivative of \(x\) with respect to \(x\) is 1. The derivative of \(\cos y\) with respect to \(x\) is \(-\sin y \cdot \frac{dy}{dx}\), using the chain rule.

Step 2: Set Up the Equation

After differentiating, we have the equation:\[1 = -\sin y \cdot \frac{dy}{dx}\]. This relation connects the derivative of \(y\) (denoted as \(\frac{dy}{dx}\)) to \(\sin y\).

Step 3: Solve for \(\frac{dy}{dx}\)

Rearrange the equation from Step 2 to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{1}{-\sin y} = -\frac{1}{\sin y}\].

Step 4: Verify Against Known Formula

In prior lessons, we learned the derivative of \(y\) with respect to \(x\) from \(x = \cos(y)\) should indeed be \(-\frac{1}{\sin y}\), considering that \(\sin y eq 0\). Thus, the implicit differentiation result agrees with the formula we previously derived.

Key Concepts

Chain Rule

The chain rule is a crucial lemma for finding derivatives of composite functions. In simpler words, it allows us to differentiate a function that is inside another function. Imagine taking apart a nested toy to see what's inside—this is how the chain rule works in differentiating layers.
In our exercise, we dealt with the equation
\[ x = \cos y \],
which is an implicit function involving a trigonometric function.
When differentiating \( \cos y \), with respect to \( x \), we can't treat\( y \)like a constant. Instead,\( y \)changes as a function of \( x \). Here's where the chain rule comes in. We apply it by:

• First, differentiating \( \cos y \) with respect to \( y \), giving us \( -\sin y \).
• Then we multiply by \( \frac{dy}{dx} \), recognizing that \( y \)is a function of \( x \).
  

Thus, the chain rule gives us \(-\sin y \cdot \frac{dy}{dx}\), which is essential in solving our original implicit differentiation problem.

Trigonometric Functions

Trigonometric functions describe relationships between the angles and sides of triangles. They are essential in modeling periodic phenomena like waves.
In our context, we are particularly interested in the cosine function, denoted by\( \cos(y) \). The cosine function is one of the primary trigonometric functions and is characterized by its periodic wave form.
Differentiating trigonometric functions, such as the cosine, brings us the sine. For instance, differentiating\( \cos(y) \)while applying implicit differentiation gives us \(-\sin(y)\). This sign change is an essential characteristic of trigonometric function derivatives.
Here's a breakdown of other useful trigonometric derivatives:

• The derivative of \( \sin(y)\)is \( \cos(y)\).
• The derivative of \( \tan(y)\)is \( \sec^2(y)\).
  

Recognizing these derivatives allows us to easily compute and verify results when engaged in calculus exercises involving trigonometric expressions.

Derivatives of Inverse Functions

Understanding derivatives of inverse functions opens doors to finding rates of change in functions expressed in reverse order. An inverse function "undoes" the effect of the original function.
For trigonometric functions, the inverse exists within specific domains, allowing us to delve into their derivatives. In this exercise,\( x = \cos(y) \)is an inverse function representation of the cosine function.
The derivative of an inverse trigonometric function usually involves a chain rule and yields expressions involving square roots or reciprocal functions. The inverse cosine, \( \cos^{-1}(x) \)or \( \text{arccos}(x) \), is among these.
Applying implicit differentiation and recognizing inverse functions lead to derivatives like:

• The derivative of \( \cos^{-1}(x) \)is \(-\frac{1}{\sqrt{1-x^2}}\).
• In the exercise, identifying \( -\frac{1}{\sin(y)} \)as the solution uses a similar approach in its implicit exploration.

This allows us to verify that our result agrees with established derivatives of inverse functions, confirming accuracy and understanding.