Question

Use the chain rule to find \(d y / d x\) for the following: (a) \(y=\left(3 x^{2}-13\right)^{3}\) (b) \(y=\left(7 x^{3}-5\right)^{9}\) (c) \(y=(a x+b)^{5}\)

Short answer

(a) \(18x(3x^2 - 13)^2\), (b) \(189x^2(7x^3 - 5)^8\), (c) \(5a(ax + b)^4\)."

Step-by-step solution

Identify Inner and Outer Functions for Part (a)

For the expression \(y=(3x^2-13)^3\), identify the inner function as \(u = 3x^2 - 13\) and the outer function as \(v = u^3\).

Differentiate Inner and Outer Functions for Part (a)

Differentiate the inner function: \(\frac{du}{dx} = \frac{d}{dx}(3x^2 - 13) = 6x\). Differentiate the outer function \(v = u^3\) with respect to \(u\): \(\frac{dv}{du} = 3u^2\).

Apply the Chain Rule for Part (a)

Using the chain rule, \(\frac{dy}{dx} = \frac{dv}{du} \times \frac{du}{dx}\). Substitute the derivatives: \(\frac{dy}{dx} = 3(3x^2 - 13)^2 \times 6x\). Simplify to obtain \(\frac{dy}{dx} = 18x(3x^2 - 13)^2\).

Repeat Steps 1-3 for Part (b)

Identify the functions for \(y = (7x^3 - 5)^9\): Inner function \(u = 7x^3 - 5\), outer \(v = u^9\). Derivative of inner: \(\frac{du}{dx} = 21x^2\), derivative of outer: \(\frac{dv}{du} = 9u^8\). Apply chain rule: \(\frac{dy}{dx} = 9(7x^3 - 5)^8 \times 21x^2\), simplifying gives \(\frac{dy}{dx} = 189x^2(7x^3 - 5)^8\).

Repeat Steps 1-3 for Part (c)

Identify the functions for \(y = (ax + b)^5\): Inner function \(u = ax + b\), outer \(v = u^5\). Derivative of inner: \(\frac{du}{dx} = a\), derivative of outer: \(\frac{dv}{du} = 5u^4\). Apply chain rule: \(\frac{dy}{dx} = 5(ax + b)^4 \times a\), simplifying gives \(\frac{dy}{dx} = 5a(ax + b)^4\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus, focusing on finding the rate at which one quantity changes with respect to another. This process is crucial for understanding how functions behave and change, allowing us to explore slopes, tangents, and instantaneous rates of change. Essentially, differentiation is about "finding the derivative."

The derivative of a function at a given point can be thought of as the slope of the tangent to the function's graph at that point. This slope indicates how steep the graph is at any given moment, providing valuable information about the function's behavior.

• For functions of the form \(y=f(x)\), the derivative is denoted \(dy/dx\).
• The process involves applying various rules, such as the power rule, the product rule, and particularly, the chain rule in cases involving composite functions.

When dealing with more complex functions, as seen in our examples, the chain rule becomes an essential tool, simplifying the process of finding derivatives of nested functions.

Inner and Outer Functions

In mathematical economics and calculus, especially when applying the chain rule, understanding the concepts of inner and outer functions is crucial. These terms help us break down composite functions, which are functions built from two or more simpler functions.

Consider a function \(y=f(g(x))\). In this scenario:

• "Inner function" refers to \(g(x)\) which is the function inside another.
• "Outer function" refers to \(f(u)\), where \(u=g(x)\).

When applying the chain rule to differentiate, it is important to:
1. Differentiate the inner function to find \(du/dx\), which means finding how \(u\) changes with respect to \(x\).
2. Differentiate the outer function with respect to \(u\), providing \(dv/du\).

This structured approach helps in systematically untangling complex expressions, simplifying the differentiation process.

Mathematical Economics

Mathematical economics involves the application of mathematical methods to represent economic theories and analyze economic problems. Differentiation, a key concept in calculus, plays a vital role in this field. By helping to find the rate at which quantities change, it aids in understanding behaviors such as:

• Maximizing profit by determining where the rate of change of cost and revenue are optimal.
• Analyzing the elasticity of demand by examining how sensitive the demand for a product is to changes in price.

Understanding how to differentiate complex functions allows economists to gain insights into these dynamic relationships.

In practice, economists use calculus to optimize functions and model economic scenarios accurately. The ability to derive and interpret derivatives, such as those calculated using the chain rule, equips them with the tools to predict changes and make informed decisions.

Calculus

Calculus is a branch of mathematics that explores change and motion, focusing on two main concepts: differentiation and integration. It has widespread application in various fields, including physics, engineering, and economics.

Differentiation, one of calculus's key techniques, enables us to find how a function changes at any point. This is especially useful for understanding movements or trends.

• Using formulas and rules, like the chain rule, allows us to handle complex functions and their derivatives with ease.
• The chain rule, specifically, aids in differentiating composite functions by breaking them down into simpler parts.

Calculus not only aids in solving real-world problems but also in developing a deeper understanding of theoretical concepts, facilitating discoveries and innovations.