Problem 308 - Calculus Volume 1

Question

For the following exercises, use implicit differentiation to find \(\frac{d y}{d x}\) $$ (x y)^{2}+3 x=y^{2} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{-2y(xy) - 3}{2x(xy) - 2y}\)

Step 1: Differentiate Both Sides

Start with the equation \((xy)^2 + 3x = y^2\). Differentiate both sides with respect to \(x\). Use the chain rule for \((xy)^2\) and implicit differentiation for \(y^2\).

Step 2: Apply the Chain and Product Rule

For the term \((xy)^2\), apply the chain rule: \(2(xy)((xy)')\), where \((xy)' = x\frac{dy}{dx} + y\) using the product rule. For \(3x\), the derivative is \(3\). For \(y^2\), using implicit differentiation, the derivative is \(2y\frac{dy}{dx}\).

Step 3: Substitute and Simplify

Substitute the derivatives into the equation: \(2(xy)(x\frac{dy}{dx} + y) + 3 = 2y\frac{dy}{dx}\). Simplify this equation.

Step 4: Solve for \(\frac{dy}{dx}\)

Collect all terms involving \(\frac{dy}{dx}\) on one side of the equation: \(2x(xy)\frac{dy}{dx} - 2y\frac{dy}{dx} = -2y(xy) - 3\). Factor out \(\frac{dy}{dx}\) on the left side, then solve for \(\frac{dy}{dx}\).

Step 5: Final Expression for \(\frac{dy}{dx}\)

Solve the equation obtained in the previous step for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-2y(xy) - 3}{2x(xy) - 2y}\).

Key Concepts

Chain Rule in Implicit Differentiation

The chain rule is a fundamental concept when it comes to differentiation, especially implicit differentiation. It helps us differentiate compositions of functions. It's crucial to make sure that we apply the chain rule correctly by identifying the inner and outer functions in a composition.
In the given exercise, we dealt with the expression \((xy)^2\). Here, recognize that \(xy\) is the inner function and squaring is the outer function.

• When you differentiate something of the form \((u)^2\), where \(u\) is a function of \(x\), the result will be \(2u \cdot u'\).

This is what the chain rule is doing for us. When combined with implicit differentiation for terms like \(y^2\), where \(y\) itself is a function of \(x\), we get a comprehensive derivative involving both \(x\) and \(y\). This is essential in solving the original exercise.

Product Rule in Implicit Differentiation

The product rule is utilized when differentiating functions where two expressions are multiplied together.It is another key tool used in implicit differentiation when differentiating expressions like \(xy\).The rule states that if you have two functions, say \(u(x)\) and \(v(x)\), their derivative is \(u'v + uv'\).

• For the exercise, \((xy)^2\) involves using both the chain rule and the product rule.
• The product rule helps differentiate \(xy\) as \(x\cdot\frac{dy}{dx} + y\).

By applying both the chain and product rules, we're able to effectively handle the composite and multipled aspects of our functions and thus manage complex implicit differentiation problems efficiently.

Differential Equations through Implicit Differentiation

Implicit differentiation is a technique used when it's difficult or impossible to solve an equation for \(y\).Such tasks often involve solving a differential equation, a type of equation that expresses a function's derivative. In the case at hand, you differentiate implicitly to find \(\frac{dy}{dx}\) since there's no explicit \(y\) expression in terms of \(x\).

• It allows us to keep the derived equation in terms of both \(x\) and \(y\). This is vital for problems like the given exercise.

After differentiating, the next step involves simplifying the derivatives and solving for \(\frac{dy}{dx}\). This process often uncovers relationships between \(y\) and \(x\) that were implicit in the original equation. Thus, differential equations and implicit differentiation work hand in hand, especially in complex algebraic functions, providing a bridge between derivatives and algebraic manipulation.