Problem 21 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find \(\frac{d y}{d x}\) if \(\left(x^{2}+y^{2}\right)^{2}=10 x y\).

Step-by-Step Solution

Answer

Question: Find the derivative of y with respect to x, given the equation \((x^2 + y^2)^2 = 10xy\). Answer: \(\frac{dy}{dx} = \frac{10y - 4x^3}{4x^2y + 4xy^2 - 10x}\)

Step 1: Rewrite the equation

Rewrite the given equation as follows: \((x^2 + y^2)^2 = 10xy\)

Step 2: Apply the implicit differentiation to both sides

We will differentiate both sides with respect to x using the chain rule: \(2(x^2 + y^2)\big(\frac{d}{d x}(x^2 + y^2)\big) = 10\big(\frac{d}{d x}(xy)\big)\)

Step 3: Differentiate each term inside parentheses

Differentiate each term inside the parentheses with respect to x and expand the equation: \(2(x^2 + y^2)(2x + 2y\frac{dy}{dx}) = 10(x\frac{dy}{dx} + y)\)

Step 4: Simplify and isolate the derivative term

Simplify the equation and isolate the \(\frac{dy}{dx}\) term: \(4(x^2 + y^2)(x + y\frac{dy}{dx}) = 10(x\frac{dy}{dx} + y)\) Now, distribute the \(4(x^2 + y^2)\) term: \(4x^3 + 4xy\frac{dy}{dx} x^2 + 4xy^2\frac{dy}{dx} = 10x\frac{dy}{dx} + 10y\) Next, group all terms with \(\frac{dy}{dx}\) on one side and simplify further: \(\frac{dy}{dx}(4x^2y + 4xy^2 - 10x) = 10y - 4x^3\)

Step 5: Solve for the derivative

Finally, divide by the polynomial to find the derivative \(\frac{dy}{dx}\) : \(\frac{dy}{dx} = \frac{10y - 4x^3}{4x^2y + 4xy^2 - 10x}\) The derivative, \(\frac{dy}{dx}\), is found as: \(\frac{dy}{dx} = \frac{10y - 4x^3}{4x^2y + 4xy^2 - 10x}\)

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation technique in calculus, especially useful when dealing with composite functions. In simpler terms, if you have a function nested inside another function, you'll likely need the chain rule to differentiate it.

For instance, consider a function of the form \((f(g(x)))\). To find its derivative, the chain rule tells us to multiply the derivative of the outer function \((f')\) evaluated at the inner function \((g)\), by the derivative of the inner function \((g')\). This can be expressed as \((f'(g(x)) \cdot g'(x))\).

• Apply the derivative of the outer function first.
• Multiply it by the derivative of the inner function.
• Ensure the inner component remains unchanged during the outer differentiation.

When dealing with implicit differentiation problems like the one in our exercise, the chain rule helps in differentiating terms that have functions of \(y\) depending on \(x\). For example, differentiating \(y^2\) with respect to \(x\) involves the chain rule: \((2y \cdot \frac{dy}{dx})\). This captures the implicit relationship between \(x\) and \(y\).

Derivative

The derivative of a function signifies its rate of change, or how it varies in response to changes in one of its variables. It forms the core of differential calculus and is used to solve problems involving change and motion.

When we differentiate a function, we're essentially finding the function's slope at any given point. The slope indicates how much the function's output \(y\) changes for a small change in the input \(x\).

• Conceptually, the derivative is the "instantaneous rate of change" of a function.
• It is the limit of the function's average rate of change over shrinking intervals.
• The derivative function \(f'(x)\) gives the slope of the original function \(f(x)\) at any \(x\).

In the problem provided, we seek \(\frac{dy}{dx}\), which is the derivative of \(y\) with respect to \(x\). This derivative tells us how \(y\) changes as \(x\) changes, reflecting their implicit relationship.

Calculus Problem Solving

Solving calculus problems effectively requires a systematic approach. First, identify the type of problem you are dealing with. Whether it's an optimization problem, a rate of change problem, or a differentiation issue, knowing this will direct the use of appropriate methods.

For implicit differentiation problems like the one in our example, follow these steps:

• Rewrite the given equation for clarity, ensuring that each function or term is prepared for differentiation.
• Differentiate both sides of the equation with respect to \(x\), using any necessary differentiation techniques.
• Simplify the resulting equation, paying attention to terms that involve \( \frac{dy}{dx} \).
• Isolate the derivative on one side of the equation to solve for \( \frac{dy}{dx} \).

When dealing with more complex equations that involve multiple terms and functions, breaking the problem into manageable parts can simplify the process and reduce errors.

Differentiation Techniques

Differentiation techniques are varied and tailored to handle a wide array of function types. Knowing which technique to use can simplify the process and lead to a correct solution.

• **Basic Rules:** These include the constant rule, power rule, and sum rule, which are essential for differentiating simpler functions.
• **Product Rule:** Used when differentiating the product of two functions, stated as \( (uv)' = u'v + uv' \).
• **Quotient Rule:** For functions in a quotient, this is expressed as \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).
• **Chain Rule:** As covered, it handles composite functions effectively.

For an implicit differentiation problem, consider:

• Differentiating directly using product and chain rules, as each term could involve multiple functions.
• Utilizing implicit differentiation, especially when \(y\) is a function of \(x\) that can't be easily isolated.

Essentially, familiarity with these techniques ensures you are prepared to handle any differentiation task, even those involving complex relationships between variables. By practicing various problems, understanding these techniques only becomes more intuitive and accessible.