Problem 15 - Calculus an Applied Approach

Question

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. Equation \(\quad\) Point \(y+x y=4 \quad(-5,-1)\)

Step-by-Step Solution

Answer

Evaluating this gives \(\frac{dy}{dx} = 0.20\) at the point \((-5, -1)\).

Step 1: Perform Implicit Differentiation

To start with, differentiate both sides of the equation with respect to \('x'\), using the chain rule and the product rule when required. The derivative of \(y\) is \(\frac{dy}{dx}\) and the derivative of \(x \cdot y\) is \(x \cdot \frac{dy}{dx} + y\), as per the product rule. So differentiating \(y + x \cdot y = 4\) gives \( \frac{dy}{dx} + x \cdot \frac{dy}{dx} + y = 0\).

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to make \(\frac{dy}{dx}\) the subject. From the equation we derived, we rewrite it as \((1 + x) \cdot \frac{dy}{dx} = -y\). Dividing both sides by \((1 + x)\) gives \(\frac{dy}{dx} = -\frac{y}{(1 + x)}.\)\)

Step 3: Substitute Given Point into The Equation

Substitute \((-5, -1)\) into \(\frac{dy}{dx} = -\frac{y}{(1 + x)}\) which gives \(\frac{dy}{dx} = -\frac{-1}{(1 + -5)}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. In essence, if you have a function nested within another function, the chain rule helps you find its derivative. This is particularly useful with implicit differentiation where functions are not explicitly solved for one variable.
When dealing with implicit differentiation, you often see variables mixed together, like in the equation provided: \( y + x y = 4 \). Here, \( y \) is implicitly defined as a function of \( x \), so both sides of the equation are differentiated with respect to \( x \). However, it's important to note that whenever \( y \) is differentiated with respect to \( x \), you apply the chain rule, resulting in \( \frac{dy}{dx} \).
In practice, you differentiate \( y \) with respect to \( x \) by noting that \( y \) is a function of \( x \) and using the chain rule, expressing changes in \( y \) through \( \frac{dy}{dx} \). Hence, the differentiation process turns out to be a systematic execution of the chain rule across each term involving \( y \).

Product Rule

The product rule comes into play when you have products of two functions to differentiate. For instance, in the expression \( x \cdot y \), both \( x \) and \( y \) are variables that depend on \( x \). The product rule states that the derivative of a product \( u \cdot v \) is \( u' \cdot v + u \cdot v' \), where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.
Applying this to \( x \cdot y \):

• Let \( u = x \), then \( u' = 1 \).
• Let \( v = y \), then \( v' = \frac{dy}{dx} \).

Using the product rule, the derivative of \( x \cdot y \) becomes \( x \cdot \frac{dy}{dx} + y \), as demonstrated. This application ensures that the differentiation of terms like \( x \cdot y \) is accurately calculated by recognizing each variable's individual contribution and applying the product rule correctly.

Derivative Evaluation

In the context of implicit differentiation, once you derive the expression for the derivative, the next step involves evaluating it at a specific point. In this exercise, the point given is \((-5, -1)\). This step is crucial because it gives the value of the derivative (or slope) of the curve described by the equation at that specific point.
To do this, substitute the \( x \) and \( y \) values into the derivative expression you found. Here, substituting \((-5, -1)\) into \( \frac{dy}{dx} = -\frac{y}{1 + x} \) simplifies to:

• Replace \( y \) with \(-1\).
• Replace \( x \) with \(-5\).

This gives \( \frac{dy}{dx} = -\frac{-1}{1 + (-5)} = \frac{1}{-4} = -\frac{1}{4} \). This result, \(-\frac{1}{4}\), is the slope of the tangent to the curve at the point \((-5, -1)\), and it encapsulates how \( y \) changes with \( x \) at that exact location.