Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(\frac{d y}{d x}\). (a) \(3 x^{2}+6 y^{2}+3 x y=10\) (b) \((x-2)^{3} \cdot(y-2)^{3}=1\) (c) \(x y^{2}+2 y=x^{2} y+1\) (d) \(\left(x^{2} y^{3}+y\right)^{2}=3 x\) (e) \(e^{x y}=y^{2}\) (f) \(x \ln \left(x y^{3}\right)=y^{2}\) (g) \(\ln (x y)=x y^{2}\)

Step-by-Step Solution

Answer

For (a), \(\frac{d y}{d x} = \frac{-3 x -3 y}{6 y +3 x}\)\nFor (b), \(\frac{d y}{d x} = \frac{-3(x-2)^{2} \cdot (y-2)^{3}}{(x-2)^{3} \cdot 3(y-2)^{2}}\)\nFor (c), \(\frac{d y}{d x} = \frac{y^{2}-2 x y}{x^{2}-2 y-x y^{2}}\)\nFor (d), \(\frac{d y}{d x} = \frac{6 x^{2} y^{2} -2 x^{2} y^{3} -2 y^{3}}{2 x^{3} y^{2} +6 x y^{3} +4 y}\)\nFor (e), \(\frac{d y}{d x} = \frac{y e^{x y}}{2 y-x y^{2}}\)\nFor (f), \(\frac{d y}{d x} = \frac{2 y x^{2} y^{2}}{x y^{2} +3 x^{2}}\)\nFor (g), \(\frac{d y}{d x} = \frac{x y +y -2 x y^{2}}{x -2 x^{2} y}\)

Step 1: Differentiate (a)

Firstly differentiate both sides of the equation with respect to x. For \(3 x^{2}+6 y^{2}+3 x y=10\), you will have \(\frac{d}{d x}( 3 x^{2}) + \frac{d}{d x}(6 y^{2}) + \frac{d}{d x}(3 x y) = \frac{d}{d x}(10)\). After differentiating and bringing terms together, your answer will be \(\frac{d y}{d x} = \frac{-3 x -3 y}{6 y +3 x}\).

Step 2: Differentiate (b)

For \((x-2)^{3} \cdot(y-2)^{3}=1\), apply the chain rule. You will get \((x-2)^{3} \cdot 3(y-2)^{2} \cdot\frac{d y}{d x} + 3(x-2)^{2} \cdot (y-2)^{3}=0\) which gives \(\frac{d y}{d x} = \frac{-3(x-2)^{2} \cdot (y-2)^{3}}{(x-2)^{3} \cdot 3(y-2)^{2}}\).

Step 3: Differentiate (c)

Similar to step 1, differentiate the equation \(x y^{2}+2 y=x^{2} y+1\) to get \(\frac{d y}{d x} = \frac{y^{2}-2 x y}{x^{2}-2 y-x y^{2}}\).

Step 4: Differentiate (d)

The complexity of \(\left(x^{2} y^{3}+y\right)^{2}=3 x\) demands the chain rule for differentiation. This yields, \(\frac{d y}{d x} = \frac{6 x^{2} y^{2} -2 x^{2} y^{3} -2 y^{3}}{2 x^{3} y^{2} +6 x y^{3} +4 y}\).

Step 5: Differentiate (e)

The equation \(e^{x y}=y^{2}\) requires logarithmic differentiation. By taking logs of both sides and differentiating, you get \(\frac{d y}{d x} = \frac{y e^{x y}}{2 y-x y^{2}}\).

Step 6: Differentiate (f)

For \(x \ln \left(x y^{3}\right)=y^{2}\), apply logarithmic differentiation to obtain \(\frac{d y}{d x} = \frac{2 y x^{2} y^{2}}{x y^{2} +3 x^{2}}\).

Step 7: Differentiate (g)

Last but not least, for \(\ln (x y)=x y^{2}\), you will differentiate to get \(\frac{d y}{d x} = \frac{x y +y -2 x y^{2}}{x -2 x^{2} y}\).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. It's based on the principle that if a variable z depends on y, which in turn depends on x, then z indirectly depends on x, and the derivative of z with respect to x can be found by multiplying the derivative of z with respect to y by the derivative of y with respect to x.

For example, when differentiating a function like \( (x-2)^{3} \cdot(y-2)^{3} = 1 \) as in the exercise, we observe two functions nested within each other. Here, we apply the chain rule to find the derivative of each part before multiplying them together. This enables us to manage more complex equations systematically and efficiently by breaking them down into simpler parts.

Applying the Chain Rule

When faced with a function like \( (x-2)^{3} \cdot(y-2)^{3} \), we treat \(y-2)\) as a function of x (through y’s dependence on x) and differentiate accordingly. The derivative then has terms involving \(\frac{d y}{d x}\), which we solve for to find the rate at which y changes with respect to x.

Logarithmic Differentiation

Logarithmic differentiation is particularly useful when dealing with functions where variables are both in the base and the exponent, such as \(e^{xy} = y^{2}\) from the exercise. This approach simplifies the differentiation process by taking the natural logarithm of both sides of the equation before differentiating.

By using the properties of logarithms, such as the product, quotient, and chain rules for logarithms, we reduce the complexity of the differentiation process. Logarithmic differentiation transforms the original equation into a form that is easier to work with and can then be differentiated using the standard rules of calculus.

Executing Logarithmic Differentiation

Consider the equation \(e^{xy} = y^{2}\). Taking the natural logarithm of both sides gives us \(xy \cdot \ln(e) = \ln(y^{2})\), then we differentiate using the product rule and the chain rule for derivatives, which results in a much more straightforward process.

Differentiate Implicit Functions

Implicit differentiation is applied to equations where y is not isolated on one side, making it impractical to solve for y before differentiating. Instead, we differentiate both sides of the equation with respect to x, treating y as an implicit function of x.

During this process, every term that contains a y is also differentiated respecting x, which introduces the derivative \(\frac{d y}{d x}\) into our equation. Solving these types of problems requires a blend of differentiation rules, including the chain rule and product rule, to manage the different forms y may appear in within an equation.

Steps in Implicit Differentiation

For instance, with the equation \(3x^{2} + 6y^{2} + 3xy = 10\), we differentiate term by term. The x-terms are straightforward, but the y-terms require us to multiply by \(\frac{d y}{d x}\) as per the chain rule. We then solve for \(\frac{d y}{d x}\) to find the desired relation between the rates of change of x and y.

Calculus Problems and Solutions

Calculus problems often range from simple direct differentiation to complex problems involving implicit functions or logarithmic differentiation. Understanding how to apply rules like the chain rule and knowing when to use techniques like logarithmic differentiation is key to solving these problems. Moreover, calculus problems provide a way to model and solve real-world phenomena, making them essential tools in fields such as physics, engineering, economics, and beyond.

Approach to Solving Problems

When approaching calculus problems and solutions, it's helpful to have a step-by-step strategy:

• Analyze the problem and identify which rules or techniques apply.
• Apply differentiation rules correctly, step by step.
• Remember to simplify where possible to make the problem easier to solve.
• Check your work for potential errors or opportunities to further simplify.

Using a structured approach can greatly assist in solving calculus problems efficiently and accurately, as shown in the step-by-step differentiation of the given exercises.