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Chapter 3: Problem 24

For the following exercises, find \(\frac{d y}{d x}\) using partial derivatives. $$ x^{2} y^{3}+\cos y=0 $$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{-2xy^3}{3x^2 y^2 - \sin y} \).

Step by step solution

01

Understand the Implicit Function

The given equation \( x^2 y^3 + \, \cos y = 0 \) is an implicit relationship between \( x \) and \( y \). We need to find the derivative \( \frac{dy}{dx} \). This involves using implicit differentiation.
02

Differentiate with respect to x

Differentiate each term of the equation \( x^2 y^3 + \cos y = 0 \) using partial derivatives. For the term \( x^2 y^3 \), apply the product rule. Differentiate \( \cos y \) partially with respect to \( y \) and multiply by \( \frac{dy}{dx} \).
03

Apply the Product Rule

For \( x^2 y^3 \), use the product rule: \( \frac{d}{dx}(x^2 y^3) = x^2 \frac{d}{dx}(y^3) + y^3 \frac{d}{dx}(x^2) \). Compute these derivatives separately. \( \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \) and \( \frac{d}{dx}(x^2) = 2x \). Thus, \( x^2 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 2x \).
04

Differentiate the Cosine Term

The derivative of \( \cos y \) with respect to \( x \) involves chain rule: \( \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \).
05

Combine Derivatives and Simplify

Combine the results from Steps 3 and 4 to obtain the full differentiated equation: \( x^2 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 2x - \sin y \cdot \frac{dy}{dx} = 0 \).
06

Solve for \( \frac{dy}{dx} \)

Factor \( \frac{dy}{dx} \) out of the terms and rearrange to solve: \( (3x^2 y^2 - \sin y) \cdot \frac{dy}{dx} = -2xy^3 \). Hence, \( \frac{dy}{dx} = \frac{-2xy^3}{3x^2 y^2 - \sin y} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a powerful technique used when dealing with equations where variables are intertwined, making it difficult or impossible to isolate one variable on one side. In the equation \( x^2 y^3 + \cos y = 0 \), both \( x \) and \( y \) are intricately related.
Since it is not possible to explicitly solve for \( y \) in terms of \( x \), implicit differentiation allows us to find the derivative \( \frac{dy}{dx} \) by assuming \( y \) as a function of \( x \).
  • First, differentiate every term of the equation with respect to \( x \).
  • Remember that when differentiating terms involving \( y \), you'll multiply the derivative by \( \frac{dy}{dx} \).
  • This helps express how changes in \( x \) affect \( y \), reflecting their implicit relationship.
Using implicit differentiation on each term enables you to maintain the balance between \( x \) and \( y \), giving you an equation you can solve for \( \frac{dy}{dx} \).
Product Rule
The product rule is essential when differentiating functions that are products of two or more functions. For the equation \( x^2 y^3 + \cos y = 0 \), the term \( x^2 y^3 \) is a product of \( x^2 \) and \( y^3 \), both depending on \( x \).The product rule states:\[\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}\]For our specific term:
  • Set \( u = x^2 \), resulting in \( \frac{du}{dx} = 2x \).
  • Set \( v = y^3 \), giving \( \frac{dv}{dx} = 3y^2 \frac{dy}{dx} \).
  • Plug these into the product rule: \( x^2 \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 2x \).
The product rule elegantly splits the function into simpler parts, ensuring that each component's rate of change is accurately captured. This separation is especially useful in functions of combined variables, like in our example.
Chain Rule
The chain rule is another fundamental technique that comes into play when a function is composed of other functions, which is useful for differentiating expressions where \( y \) is a function of \( x \). In our equation, the term \( \cos y \) requires the use of the chain rule.The chain rule can be expressed as:\[\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)\]For \( \cos y \), consider:
  • The outer function \( f(y) = \cos y \) gives \( f'(y) = -\sin y \).
  • The inner function is \( y = g(x) \), with \( g'(x) = \frac{dy}{dx} \).
  • Applying the chain rule: \( \frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx} \).
By using the chain rule, you can handle nested or composite functions, allowing you to find how changes in \( x \) influence the overall function. This is critical when differentiating expressions in implicit relations between variables.

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