Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ g(x)=x^{4}+x^{2} $$

Step-by-Step Solution

Answer

The derivative of \( g(x) = x^4 + x^2 \) is \( 4x^3 + 2x \).

Step 1: Identify the Function and Rewrite the Expression

We are given the function \( g(x) = x^4 + x^2 \). We need to find its derivative at \( x \) using the definition of a derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Let's replace \( f(x) \) with \( g(x) \) in this formula.

Step 2: Substitute and Expand \( g(x+h) \)

Substitute \( g(x+h) = (x+h)^4 + (x+h)^2 \) into the derivative formula. We need to expand this expression: \( (x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \) and \( (x+h)^2 = x^2 + 2xh + h^2 \).

Step 3: Plug Expanded Expressions into Derivative Definition

Substituting these expansions back into the original derivative formula we get: \[\frac{(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 + x^2 + 2xh + h^2) - (x^4 + x^2)}{h}\]This simplifies to: \[\frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4 + 2xh + h^2}{h}\]

Step 4: Simplify the Expression by Cancelling \( h \)

Factor \( h \) out of the numerator: \[\frac{h(4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h)}{h}\]Cancelling \( h \) from the numerator and denominator gives: \[ 4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h \].

Step 5: Take the Limit as \( h \to 0 \)

Applying the limit \( h \to 0 \) to the simplified expression:\[\lim_{h \to 0} (4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h) = 4x^3 + 2x\]The terms involving \( h \) go to zero because they are multiplied by \( h \) or its higher powers.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It focuses on the concepts of limits, derivatives, integrals, and infinite series. At its core, calculus provides a framework for understanding the dynamics of changing quantities.
Derivatives and integrals are the two main tools in calculus. While derivatives are used to measure the rate of change (how much one variable changes with respect to another), integrals compute the accumulation of quantities, like areas under curves.
In this discussion, we are focusing on derivatives, which allow us to understand the behavior of functions by observing their rate of change at any given point.

• Example: Calculating the speed of a car over an interval of time involves understanding its derivative.
• Applications: Used extensively in science, engineering, and economics for modeling and problem-solving.

Mastering calculus opens up a vast toolkit for solving real-world problems relating to rates and accumulations.

Limit Definition

The concept of limits forms the foundation of calculus. It's crucial for defining derivatives and integrals. A limit provides a way to understand what value a function approaches as the input gets closer to a certain point.
When we define the derivative of a function at a point using a limit, we use the expression:\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
This formula represents the slope of the tangent line to the function at that point. The limit ensures that we find an instantaneous rate of change by making the interval infinitesimally small.

• Understanding: By considering the change in outputs relative to inputs as the difference approaches zero, we can precisely determine moments perfectly.
• Application: Such concepts are used in physics to determine the velocity of an object at a particular moment in time.

Grasping this definition deeply helps in understanding how derivatives work under the hood.

Polynomial Functions

Polynomial functions are mathematical expressions involving sums of powers of variables. They are among the simplest types of functions and appear frequently in calculus problems.
A polynomial function is of the form:
\[ P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \]
where \(a_n, a_{n-1}, \ldots, a_0\) are constants. In our example, the function
\(g(x) = x^4 + x^2\) is a polynomial of degree 4.

• Characteristics: Polynomial functions can have varying degrees; the degree refers to the highest power of \(x\).
• Ease of Differentiation: Differentiating polynomial functions is straightforward, as it follows the power rule which simplifies calculation.
• Graphing: They tend to have smooth, continuous curves which makes them easy to study visually.

Understanding how to work with polynomial functions is essential for solving a variety of calculus problems involving derivatives.

Differentiation Techniques

In calculus, differentiation is the process of finding a derivative. It transforms a function into another function, representing an instantaneous rate of change. Several techniques exist for differentiation, each useful in different contexts.
The primary technique used in our exercise is the limit definition. However, several rules simplify differentiation, especially of polynomial functions:

• Power Rule: For any term \(x^n\), the derivative is given by \(nx^{n-1}\). This was used here because each term in a polynomial is a power of \(x\).
• Sum Rule: The derivative of a sum of functions is simply the sum of their derivatives.
• Product and Quotient Rules: Useful when dealing with products or quotients of functions but not necessary for simple polynomials.

Using these techniques allows us to quickly and accurately find derivatives without always reverting to limit definitions, making problem-solving more efficient.