Problem 306 - Calculus Volume 1

Question

Use implicit differentiation to find \(\frac{d y}{d x}\). $$ -x y-2=\frac{x}{7} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{-1/7 - y}{x}.\)

Step 1: Identify Implicit Equation

The given implicit equation is \(-xy - 2 = \frac{x}{7}\). This is a combination of different variables and needs implicit differentiation.

Step 2: Differentiate Both Sides

We apply differentiation to both sides of the equation with respect to \(x\). The derivative of the left side \(-xy-2\) with respect to \(x\) is \(\frac{d}{dx}(-xy) - \frac{d}{dx}(2)\). For the right side, \(\frac{d}{dx}\left(\frac{x}{7}\right) = \frac{1}{7}\).

Step 3: Apply the Product Rule

For the term \(-xy\) on the left side, use the product rule: if \(u(x) = x\) and \(v(y) = y\), then \(\frac{d}{dx}(uv) = u'v + uv'\). So, \(\frac{d}{dx}(-xy) = -\left(y + x\frac{dy}{dx}\right)\).

Step 4: Combine Derivatives

Now combine the derivatives: \(-\left(y + x\frac{d y}{d x}\right) - 0 = \frac{1}{7}\).

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

Simplify and solve for \(\frac{dy}{dx}\): \(-y - x\frac{dy}{dx} = \frac{1}{7}. \) Rearrange to get \(-x\frac{dy}{dx} = \frac{1}{7} + y. \)Divide both sides by \(-x\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-1/7 - y}{x}.\)

Key Concepts

Product Rule

When we see expressions like \(-xy\), involving the multiplication of two different functions, the product rule of differentiation becomes an essential tool. The product rule is used to find the derivative of the product of two functions. If we have two differentiable functions, \(u(x)\) and \(v(x)\), then the product rule is given by the formula: \[\frac{d}{dx}(uv) = u'v + uv'\].This rule tells us that to differentiate a product of two functions, we first differentiate the first function and multiply it by the second one, then add it to the product of the first function and the derivative of the second function.

• In the exercise, \(u(x) = x\) and \(v(y) = -y\).
• Thus, applying the product rule, we differentiate to get \(- (y + x\frac{dy}{dx})\).
• This helps us handle functions where both variables multiply each other directly.

Differentiation

Differentiation is a fundamental concept of calculus, and it refers to the process of finding the derivative of a function. The derivative represents the rate of change or the slope of the curve at any point. When differentiating implicitly, as seen in the exercise, we work with equations where the variables are intertwined and not expressed explicitly.

• Differentiating both sides of an equation involves applying derivatives to each term.
• On the left side, derivative of \(-xy - 2\) with respect to \(x\), which used the product rule, involves treating \(y\) as a function of \(x\).
• On the right side, differentiation of \(\frac{x}{7}\) is straightforward, yielding \(\frac{1}{7}\).

Putting these steps together allows us to account for both direct and implicit relationships between variables.

Implicit Functions

In calculus, implicit functions refer to functions that are not stated explicitly as \((y = f(x))\), rather they involve a relation between \(x\) and \(y\) that defines \((y)\) implicitly. Implicit differentiation is a technique used to find derivatives when dealing with such relationships. Implicit functions require us to differentiate both sides of the equation with respect to \(x\) while considering \(y\) as a function of \(x\). In the exercise given:

• The equation \(-xy - 2 = \frac{x}{7}\) is not solved for \(y\). Instead, it shows how \(x\) and \(y\) together satisfy a condition.
• Implicit differentiation allows us to determine \(\frac{dy}{dx}\) when these variables are mixed.
• After differentiating and simplifying, you rearrange to solve for \(\frac{dy}{dx}\), eventually leading to \((\frac{dy}{dx} = \frac{-1/7 - y}{x})\).

This technique is especially useful when explicit separation of variables is difficult or impossible.