Question

In Exercises, find \(d y / d x\) implicitly. $$ \ln x y+5 x=30 $$

Short answer

The derivative of the function \(y\) with respect to \(x\) (that is, \(dy/dx\)) is \(y′ = -5y(x - 1/y)\).

Step-by-step solution

Apply logarithmic differentiation to split the term inside the natural log

Take the natural log on both sides of the equation: \(\ln(xy) = 30 - 5x\). Applying the log property \(\ln(a \cdot b) = \ln(a) + \ln(b)\), splits the term inside the log on the left-hand side resulting in: \(\ln(x) + \ln(y) = 30 - 5x\).

Differentiate both sides of the equation with respect to \(x\)

Differentiate each side of the equation with respect to \(x\): \[\frac{d}{dx}(\ln(x) + \ln(y)) = \frac{d}{dx}(30 - 5x).\] Using the chain rule on the left side results in: \(1/x + \frac{1}{y} \cdot y′ = -5\).

Solve for \(y'\)

Rearrange the derived equation to solve for \(y'\): \(y′ = -5y(x - 1/y)\).

Key concepts

Logarithmic Differentiation

Logarithmic differentiation is a clever technique utilized in calculus. It helps to simplify the process of finding derivatives of products, quotients, or powers of functions. This method involves taking the logarithm of each side of an equation before differentiating, usually aimed at breaking down complex expressions.Here’s why it's useful:

• Reduces complexity by turning multiplication into addition or division into subtraction.
• Makes it easier to handle functions with variables raised to powers.
• Allows for implicit differentiation, providing further simplification in some cases.

In situations where a function is an implicit relation, like our exercise: \(\ln(xy) + 5x = 30\), applying logarithmic differentiation can transform \(\ln(xy)\) into \(\ln(x) + \ln(y)\), easing the pathway to find the derivative, \(dy/dx\). It splits compound terms for easier management when derivatives are taken.

Natural Logarithm

The natural logarithm, commonly represented as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is approximately 2.718. In calculus, it plays a crucial role due to its elegant derivative properties and its ability to simplify differentiation processes through its logarithmic identities.Natural logarithms have special properties that make them particularly useful:

• They transform multiplicative relationships into additive ones.
• The derivative of \(\ln(x)\) is simple: \(\frac{1}{x}\).
• Helps break down complex algebraic expressions, simplifying implicit differentiation.

In our exercise, recognizing that \(\ln(xy)\) equals \(\ln(x) + \ln(y)\) allows us to simplify the equation for easier differentiation. This breakdown is a foundational concept in handling functions within logarithms effectively, aligning perfectly with calculus needs.

Chain Rule

The chain rule is a fundamental aspect of differentiation in calculus that enables the differentiation of composite functions. It acts as a bridge to finding the derivative of a function inside another function effectively.How the chain rule comes into play:

• Evaluate the outer function's derivative while maintaining the inner function as is.
• Multiply by the derivative of the inner function.

In our step-by-step solution, the expression \(\ln(y)\) involves \(y\), which is itself a function of \(x\). Thus, differentiating \(\ln(y)\) requires the chain rule:\(\frac{d}{dx}(\ln(y)) = \frac{1}{y} \cdot \frac{dy}{dx}\).This way, you're able to incorporate both the derivative of the natural log and the derivative of \(y\) concerning \(x\), ensuring that all implicit relationships are taken into account.

Derivative

The concept of a derivative is central to calculus, representing the rate at which a function changes at any given point. With respect to a function, the derivative tells us the slope of the tangent line at a particular point on a curve.Several reasons why derivatives are important:

• They reveal rates of change, making them crucial in motion, growth, and decay scenarios.
• Interface with various rules like product rule, quotient rule, and chain rule for flexible applications.
• Essentials in solving optimization problems, finding minimums and maximums of functions.

In our examined exercise, finding the derivative \(dy/dx\) required using implicit differentiation due to the involvement of \(y\) within the equation itself. Implicit differentiation allows us to handle derivatives involving y as a function of \(x\), providing a comprehensive method for tackling such equations head-on.