Problem 12 - University Calculus: Early Transcendentals

Question

Use implicit differentiation to find \(d y / d x\). $$x^{4}+\sin y=x^{3} y^{2}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{3x^2 y^2 - 4x^3}{\cos y - 2x^3 y}\)

Step 1: Differentiate both sides with respect to x

Starting with the equation \(x^4 + \sin y = x^3 y^2\), differentiate both sides with respect to \(x\). The left side will become \(4x^3 + \cos y \frac{dy}{dx}\). For the right side, use the product rule: \(\frac{d}{dx}(x^3 y^2) = x^3 \cdot \frac{d}{dx}(y^2) + y^2 \cdot \frac{d}{dx}(x^3)\), which simplifies to \(x^3 (2y \frac{dy}{dx}) + y^2 \cdot 3x^2\).

Step 2: Simplify the differentiated equation

Now substitute the derivatives into the equation, we have: \(4x^3 + \cos y \frac{dy}{dx} = x^3 (2y \frac{dy}{dx}) + 3x^2 y^2\).

Step 3: Isolate terms involving dy/dx

Collect all terms involving \(\frac{dy}{dx}\) on one side: \(\cos y \frac{dy}{dx} - 2x^3 y \frac{dy}{dx} = 3x^2 y^2 - 4x^3\).

Step 4: Factor out dy/dx

Factor \(\frac{dy}{dx}\) from the left side: \(\frac{dy}{dx} (\cos y - 2x^3 y) = 3x^2 y^2 - 4x^3\).

Step 5: Solve for dy/dx

Solve for \(\frac{dy}{dx}\) by dividing both sides by \( (\cos y - 2x^3 y) \): \(\frac{dy}{dx} = \frac{3x^2 y^2 - 4x^3}{\cos y - 2x^3 y}\).

Key Concepts

Product Rule

The product rule is a fundamental derivative rule essential for differentiating the product of two functions. It states that when you have a product of two functions, say \( f(x) \) and \( g(x) \), the derivative can be found using:

• \((fg)' = f'g + fg'\)

This means if you want to find the derivative of the product \( x^3 y^2 \) with respect to \( x \), you apply the product rule.
The steps in application are:

• Differentiate \( x^3 \) to get \( 3x^2 \).
• Keep \( y^2 \) constant and multiply the result by \( 3x^2 \).
• Next, differentiate \( y^2 \), treating \( y \) as a function of \( x \), which yields \( 2y \frac{dy}{dx} \).
• Multiply this by \( x^3 \).

Finally, sum these two products to get the derivative: \[ \frac{d}{dx}(x^3 y^2) = 3x^2 y^2 + x^3 (2y \frac{dy}{dx}) \] This shows the dual nature of differentiation when it comes to products.

Derivatives

Derivatives are central to calculus. They represent the rate at which a function is changing at any given point. In the context of implicit differentiation, derivatives help in finding the rate at which one variable changes with respect to another when they are intertwined in an equation.
Here, in the exercise, we apply derivative operations to both sides of the equation \(x^4 + \sin y = x^3 y^2\). This requires differentiating each term individually.

• The power rule simplifies \(x^4\) to \(4x^3\).
• For \(\sin y\), introduce \(\cos y \frac{dy}{dx}\) because the derivative of \(\sin y\) involves the chain rule.

Implicit differentiation is crucial when dealing with equations where \(x\) and \(y\) are not explicitly separated, making it possible to find \(\frac{dy}{dx}\) without having an explicit expression for \(y\) in terms of \(x\).
These derivative operations enable us to transform a complex equation into a solvable format for \(\frac{dy}{dx}\).

Trigonometric Functions

Trigonometric functions, such as \( \sin \theta \), \( \cos \theta \), and necessary others, are vital in calculus for modeling situations that involve periodic or wave-like phenomena. When working with these functions in differentiation, you often encounter derivatives like:

• The derivative of \( \sin y \) is \( \cos y \frac{dy}{dx} \).
• This follows the chain rule, where you differentiate the outer function (\( \sin \)) first and then multiply by the derivative of the inside function (\( y \)).

In implicit differentiation, recognizing how to differentiate these functions becomes crucial as they don't always appear in straightforward expressions.
The trigonometric component extends the use of implicit differentiation in complex equations by adding periodic variables that interact with straight-line variables, creating multi-dimensional understanding in function behaviors.