Problem 238 - Calculus Volume 3

Question

Find \(\frac{d y}{d x}\) using partial derivatives. \(x^{2} y^{3}+\cos y=0\)

Step-by-Step Solution

Answer

Differentiate using implicit differentiation.

Step 1: Recognize Implicit Differentiation Requirement

The given equation \(x^2 y^3 + \cos y = 0\) includes both variables \(x\) and \(y\) intertwined in such a way that \(y\) is dependent on \(x\). This requires implicitly differentiating the entire equation with respect to \(x\).

Step 2: Differentiate Each Term

Differentiate each term of the equation \(x^2 y^3 + \cos y = 0\) with respect to \(x\):- The term \(x^2 y^3\) can be differentiated using the product rule, treating \(x^2\) as one function and \(y^3\) as another. This gives us \(\frac{d}{dx}(x^2 y^3) = x^2 \frac{d}{dx}(y^3) + y^3 \frac{d}{dx}(x^2)\).- The term \(\cos y\) is differentiated as \(-\sin y \cdot \frac{dy}{dx}\).

Step 3: Apply the Product Rule

Apply the product rule to differentiate \(x^2 y^3\):\[ x^2 \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 \cdot 2x = 0 \]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. These derivatives allow us to see how a function changes as one specific variable changes, while keeping other variables constant.

In the context of our problem, we have an implicit function involving two variables, \(x\) and \(y\). To solve for the derivative of \(y\) with respect to \(x\), we need to treat \(y\) as a dependent variable influenced by \(x\). This leads to taking partial derivatives to focus on one aspect of change at a time.

• Partial derivatives break down the change in a multi-variable function to single variable components.
• They are crucial for functions where we can't solve explicitly for one of the variables.
• They help us in navigating through problems involving implicit differentiation by isolating the contribution of one variable's change to the overall system.

Product Rule

The product rule is an essential tool in calculus used when differentiating products of two functions. This is particularly useful in implicit differentiation problems like the one given, involving terms like \(x^2 y^3\).

To apply the product rule, remember that if you have two functions \(u(x)\) and \(v(x)\), their derivative is given by:

• \(\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)\)

In our equation \(x^2 y^3\), we let \(u(x) = x^2\) and \(v(x) = y^3\). This helps compute the derivative as:

• \(x^2 \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 \cdot 2x\)

This highlights the step-by-step application of the product rule, allowing us to tackle implicit differentiation effectively.

Dependent Variable

A dependent variable is a variable whose value depends on one or more other variables, known as independent variables. In the equation \(x^2 y^3 + \cos y = 0\), \(y\) is the dependent variable because its value relies on the value of \(x\).

Understanding dependent variables is crucial in implicit differentiation because:

• We must interpret \(y\) as dependent on \(x\), not as an independent static entity.
• This builds the basis for taking derivatives, as changes in \(x\) will influence \(y\).
• Implicit differentiation leverages this dependency to find derivatives of expressions where \(y\) isn't solely on one side of the equation.

By acknowledging \(y\) as a dependent variable, we set the stage for applying the chain rule and other differentiation techniques effectively.

Differentiating Equations

Differentiating equations involves applying calculus rules to find derivatives. In problems where variables are intertwined, implicit differentiation is the technique used. This helps differentiate equations that imply a relationship between the dependent and independent variables.

Steps for differentiating implicit equations include:

• Identify which variable is dependent, often \(y\) in terms of \(x\).
• Apply standard rules of differentiation such as the chain rule, product rule, and the understanding of trigonometric derivatives.
• After differentiating, substitute derivations back into the equation to solve for the required derivative, \(\frac{dy}{dx}\).

This approach ensures that we respect the dependency structure between variables and allows us to smoothly find derivatives even when functions cannot be easily separated.