Problem 42 - Single Variable Calculus

Question

Use implicit differentiation to find \(d y / d x\). $$ \ln x y-y^{2}=5 $$

Step-by-Step Solution

Answer

The derivative, using implicit differentiation, is given by \(y' = \frac{dy}{dx} = \frac{-y}{x - 2xy}\).

Step 1: Differentiate both sides with respect to x

We will differentiate both sides of the given equation with respect to x, treating y as a function of x. \( \frac{d}{dx} (\ln(xy) - y^2) = \frac{d}{dx} 5\)

Step 2: Apply chain rule

For the left-hand side of the equation, apply the chain rule: \(\frac{d}{dx} (\ln(xy)) - \frac{d}{dx} (y^2) = 0\) Since the right-hand side of the equation is a constant, its derivative is 0.

Step 3: Differentiate terms separately

Differentiate each term separately. Recall that \( \frac{d}{dx}(\ln u) = \frac{u'}{u}\) for a differentiable function u. \( \frac{xy' + x'y}{xy} - 2yy' = 0\)

Step 4: Simplify and solve for dy/dx

Now we'll simplify the equation and solve for \(dy/dx\) or y': \( \frac{y+xy'}{x} - 2yy' = 0\) First, multiply both sides by x to eliminate the fraction: \(y + xy' - 2xyy' = 0\) Now, isolate the \(y'\) term: \(y' (x - 2xy) = -y\) Finally, divide by \((x - 2xy)\) to find the derivative: \(y' = \frac{dy}{dx} = \frac{-y}{x - 2xy}\)

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for finding the derivative of a composite function. It allows us to differentiate functions within functions—think of it as peering into layers. When differentiating a function like \( \ln(xy) \), which includes a multiplication inside the parentheses, we apply the chain rule to break it into manageable parts. This means we first differentiate the outer function and then multiply it by the derivative of the inner function.

• For \( \ln(xy) \), let \( u = xy \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).
• Then find the derivative of \( u = xy \) with respect to \( x \), which gives \( y + xy' \) by the product rule.

So, combining these, \( \frac{d}{dx} \ln(xy) \) becomes \( \frac{y + xy'}{xy} \). This simplification helps in solving implicit differentiation problems.

Derivative

In calculus, a derivative represents the rate of change of a function with respect to a variable. It provides us with the slope of the function at any point. For implicit differentiation, we consider \( y \) as a function of \( x \) even if it's not explicitly solved.

• Here, we assume \( y = f(x) \), so every time we derive \( y \), we multiply by \( y' \), the derivative of \( y \) with respect to \( x \).
• For example, the derivative of \( y^2 \) is \( 2yy' \).

Implicit differentiation helps us solve derivatives when functions are intertwined, allowing us to find \( \frac{dy}{dx} \). In this exercise, solving for \( y' \) gives the change of \( y \) in terms of \( x \).

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \), where \( e \approx 2.718 \). It's often used in calculus because of its natural properties—especially with exponential growth and decay models.

• For the function \( \ln(xy) \), applying differentiation involves using properties of logarithms and the chain rule.
• Remember, \( \ln(ab) = \ln(a) + \ln(b) \). This property can sometimes simplify differentiation.

In our problem, understanding how to differentiate logarithmic functions assists in separating complex expressions into simpler parts, essential for solving implicit equations.

Functions of Multiple Variables

When dealing with functions of multiple variables, like \( xy \) in our exercise, it’s crucial to understand how each variable independently affects the function. This often involves partial derivatives or techniques that address how one variable influences another.

• In \( \ln(xy) - y^2 = 5 \), \( x \) and \( y \) are interdependent, creating a system that needs implicit differentiation.
• An implicit function theory comes into play because \( y \) is not isolated but connected to \( x \).

By treating one variable as a constant while differentiating with respect to another, we can unravel the relationships within these functions. These skills are fundamental in fields that require modeling real-world systems.