Problem 230 - Calculus 3

Question

For the following exerises, find \(\frac{d y}{d x}\) using parial derivatives. $$ \sin (6 x)+\tan (8 y)+5=0 $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -\frac{3\cos(6x)}{4\sec^2(8y)} \).

Step 1: Understand the Implicit Function

The function given is \( \sin(6x) + \tan(8y) + 5 = 0 \). Here, \( y \) is implicitly defined as a function of \( x \). We need to find \( \frac{dy}{dx} \) using partial derivatives. This involves implicit differentiation.

Step 2: Differentiate both sides with respect to x

Differentiate the entire equation with respect to \( x \). Remember, when differentiating terms involving \( y \), you must include the derivative \( \frac{dy}{dx} \) using the chain rule.- Differentiate \( \sin(6x) \) with respect to \( x \): \( \frac{d}{dx}[\sin(6x)] = 6 \cos(6x) \).- Differentiate \( \tan(8y) \) using the chain rule: \( \frac{d}{dx}[\tan(8y)] = 8 \sec^2(8y) \cdot \frac{dy}{dx} \).- The constant 5 differentiates to 0.Together, this gives:\[ 6\cos(6x) + 8\sec^2(8y) \cdot \frac{dy}{dx} = 0 \]

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

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \): \[ 8\sec^2(8y) \cdot \frac{dy}{dx} = -6\cos(6x) \] Now, divide both sides by \( 8\sec^2(8y) \): \[ \frac{dy}{dx} = -\frac{6\cos(6x)}{8\sec^2(8y)} \] This simplifies further to: \[ \frac{dy}{dx} = -\frac{3\cos(6x)}{4\sec^2(8y)} \]

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique used when dealing with equations where the dependent variable is not isolated on one side. When given an equation like \( \sin(6x) + \tan(8y) + 5 = 0 \), the relationship between \( x \) and \( y \) is not explicitly solved for \( y \), which means we cannot simply take the derivative of \( y \) as a function of \( x \). Instead, we treat \( y \) as an implicit function of \( x \), and differentiate each term with respect to \( x \).

• The goal of implicit differentiation is to find the derivative \( \frac{dy}{dx} \).
• This involves knowing how to apply differentiation rules to multiple terms involving both \( x \) and \( y \).

Through implicit differentiation, we are able to derive a relationship that yields the required derivative even without having \( y \) explicitly expressed as a function of \( x \). This is particularly useful in complex equations like the one given in the exercise.

Chain Rule

The chain rule is an essential differentiation tool that applies when calculating the derivative of a composite function. Whenever a function is applied to another function, the chain rule helps us differentiate these scenarios. In our problem:

• We have \( \sin(6x) \), which is a simple chain where \( 6x \) is the inner function. Differentiating it gives \( 6\cos(6x) \).
• In the case of \( \tan(8y) \), \( 8y \) is the inner function, and it involves \( y \), meaning we must multiply by \( \frac{dy}{dx} \). The chain rule yields \( 8\sec^2(8y)\cdot \frac{dy}{dx} \).

Whenever the derivative of an inner function \( f(x) \) is multiplied with the derivative of the main function \( f(g(x)) \), the chain rule is applied. It is crucial for working with nested functions and combos of variables like in the exercise provided.

Trigonometric Functions

Trigonometric functions like sine and tangent often appear in differentiation problems. They require special attention because of their unique properties and derivatives.

• The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \), and this is critical in differentiating terms such as \( \sin(6x) \).
• Derivative of \( \tan(u) \) is \( \sec^2(u) \), which is evident in differentiating \( \tan(8y) \).

By understanding these basic trigonometric derivatives, along with the application of the chain rule, we simplify finding derivatives involving trigonometric functions, like those in our equation. Be mindful of the chain inside functions using these trigonometric identities.

Derivative Calculation

The final goal in problems like this is the calculation of \( \frac{dy}{dx} \), the derivative that denotes the rate of change of \( y \) with respect to \( x \).

• First, each part of the equation is differentiated with respect to \( x \), keeping in mind the implicit relationship of \( y \) to \( x \).
• The differentiated equation, \( 6\cos(6x) + 8\sec^2(8y)\cdot\frac{dy}{dx} = 0 \), shows the combination of partial derivatives.
• Solving for \( \frac{dy}{dx} \) involves isolating it by moving other terms, and then dividing.

This process results in the final derivative: \( \frac{dy}{dx} = -\frac{3\cos(6x)}{4\sec^2(8y)} \), achieved by understanding implicitly defined functions and adeptly using calculus rules. This approach to derivative calculation reveals deeper relationships within the equation's structure.