Problem 6 - Calculus Early Transcendentals

Question

Find \(d y / d x\) by implicit differentiation. \(x^{3} y^{2}-5 x^{2} y+x=1\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-3x^2 y^2 + 10xy - 1}{2x^3 y - 5x^2} \)

Step 1: Differentiate both sides with respect to x

Start by differentiating each term of the given equation implicitly with respect to \( x \). For computating implicit differentiation of terms involving both \( x \) and \( y \), apply the product rule and chain rule.- The derivative of \( x^3 y^2 \) is \( 3x^2 y^2 + x^3 \cdot 2y \cdot \frac{dy}{dx} \), using the product rule and the chain rule.- The derivative of \(-5x^2 y\) is \(-10xy - 5x^2 \cdot \frac{dy}{dx} \), again using the product rule and chain rule.- The derivative of \( x \) with respect to \( x \) is 1.- The derivative of the constant 1 is 0.Combining these, the differentiation gives:\[ 3x^2 y^2 + 2x^3 y \frac{dy}{dx} - 10xy - 5x^2 \frac{dy}{dx} + 1 = 0 \]

Step 2: Collect terms with dy/dx

Now we need to collect all the terms that involve \( \frac{dy}{dx} \) to one side of the equation. Start by rearranging your differentiated equation:\[ 2x^3 y \frac{dy}{dx} - 5x^2 \frac{dy}{dx} = -3x^2 y^2 + 10xy - 1 \]

Step 3: Factor out dy/dx

Factor out \( \frac{dy}{dx} \) from the left side of the equation:\[ \frac{dy}{dx} (2x^3 y - 5x^2) = -3x^2 y^2 + 10xy - 1 \]

Step 4: Solve for dy/dx

Finally, solve for \( \frac{dy}{dx} \) by dividing both sides of the equation by \( (2x^3 y - 5x^2) \):\[ \frac{dy}{dx} = \frac{-3x^2 y^2 + 10xy - 1}{2x^3 y - 5x^2} \]

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus when you need to differentiate expressions involving products of two functions. It allows us to tackle complex problems like differentiating the term \( x^3 y^2 \) in our original exercise.For two functions, \( u(x) \) and \( v(x) \), the product rule states:

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

In our problem, consider \( u = x^3 \) and \( v = y^2 \). Applying the product rule:- Differentiate \( u(x) = x^3 \) to get \( u'(x) = 3x^2 \).- Differentiate \( v(x) = y^2 \) with respect to \( x \) using the chain rule, giving \( 2y \cdot \frac{dy}{dx} \).Substituting, we have:

• \( u'(x)v(x) = 3x^2 \cdot y^2 \)
• \( u(x)v'(x) = x^3 \cdot 2y \cdot \frac{dy}{dx} \)

Combine these results to get the derivative of \( x^3y^2 \) as part of implicit differentiation for our equation.

Chain Rule

The chain rule is an essential calculus principle used to differentiate composite functions, such as in our given problem when differentiating terms that include both \( x \) and \( y \).The chain rule states:

• If a function \( y = f(g(x)) \), then its derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This rule is especially useful in implicit differentiation where one variable is expressed in terms of another implicitly.In our example, consider differentiating \( y^2 \) with respect to \( x \):- Here, \( y \) is a function of \( x \), so differentiate \( y^2 \) by applying the chain rule: \( 2y \cdot \frac{dy}{dx} \).Such use of the chain rule allows us to effectively handle terms where \( y \) is not explicitly a function of \( x \), a common occurrence in implicit differentiation problems.

Calculus Problem Solving

Solving calculus problems often requires implementing strategies step-by-step, which is vital for solving the implicit differentiation problem here.Begin by differentiating each term:- Differentiate with respect to \( x \), applying both the product and chain rules where necessary.- For example, differentiate \( x^3y^2 \) and \(-5x^2y\) using the product and chain rules.Then, follow these steps:

• Collect all terms containing \( \frac{dy}{dx} \) on one side to isolate it.
• Factor \( \frac{dy}{dx} \) out from these terms.
• Solve for \( \frac{dy}{dx} \) to find the derivative, using algebra to rearrange the equation if needed.

Precision and attention to detail in each step are crucial. When terms involve both \( x \) and \( y \), accurate application of rules simplify the process, leading to the final solution for \( \frac{dy}{dx} \) as seen in our example problem.