Problem 10 - 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 \(x\). $$ f(x)=x^{4} $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = x^4 \) is \( 4x^3 \).

Step 1: Identify the Function and Variable

The given function is \( f(x) = x^4 \). We need to find the derivative of this function with respect to \( x \) at any point \( x \).

Step 2: Set Up the Difference Quotient

The difference quotient for the derivative of \( f(x) \) is given by \( \frac{f(x+h) - f(x)}{h} \). First, calculate \( f(x+h) \), which is \((x+h)^4\).

Step 3: Expand \((x+h)^4\) Using the Binomial Theorem

Expand \((x+h)^4\) using the binomial theorem: \[(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4.\]

Step 4: Substitute and Simplify the Difference Quotient

Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient:\[\frac{(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4}{h} = \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h}.\]Cancel \( x^4 \) and then simplify by dividing each term by \( h \):\[4x^3 + 6x^2h + 4xh^2 + h^3.\]

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

Calculate the limit of the simplified expression as \( h \to 0 \):\[\lim_{h \to 0} (4x^3 + 6x^2h + 4xh^2 + h^3) = 4x^3.\]Since the terms containing \( h \) vanish as \( h \to 0 \), we are left with \( 4x^3 \).

Step 6: Conclude the Derivative

We have found the derivative of the function \( f(x)=x^4 \) to be \( f'(x) = 4x^3 \).

Key Concepts

Difference Quotient

A fundamental concept in calculus is the difference quotient, which is crucial for finding derivatives. It essentially measures how a function changes as its input changes by a tiny amount. This concept forms the backbone of derivative calculation. For any given function \( f(x) \), the difference quotient is given by:

• \( \frac{f(x+h) - f(x)}{h} \)

Here, \( h \) is a small, non-zero number representing a tiny increment from \( x \). The numerator \( f(x+h) - f(x) \) represents the change in function value, and dividing by \( h \) scales this change to reflect a rate of change. To find a derivative, we approach the situation where \( h \) becomes infinitesimally small, which leads us to the next concept: limits. Once simplified and evaluated as \( h \) approaches zero, this expression gives us the instantaneous rate of change, or the derivative, of the function at that point.

Binomial Theorem

The binomial theorem is a powerful tool that helps expand expressions raised to a power. In our exercise, we had \( (x+h)^4 \), which requires expansion to simplify the difference quotient.
The binomial theorem states that for any non-negative integer \( n \):

• \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\)

This formula allows us to break down the power of a sum into a series of terms. Each term involves binomial coefficients \( \binom{n}{k} \), which can be calculated using "n choose k," referring to the number of ways to choose k elements from a total of n without regard to order.
In our function, expanding \( (x+h)^4 \) gives us a polynomial:

• \( x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \)

This breakdown allows us to carefully manipulate each term, preparing it for substitution back into the difference quotient.

Limit Process

The limit process is the final and critical step in finding the derivative using the difference quotient. When we calculate \(\lim_{h \to 0} \left(\frac{f(x+h) - f(x)}{h}\right),\)we evaluate how the expression behaves as \( h \) gets closer and closer to zero. This procedure is what we call finding the limit.
For our function \( f(x) = x^4 \), after expanding and simplifying the difference quotient, we had:

• \( 4x^3 + 6x^2h + 4xh^2 + h^3 \)

All terms that include \( h \) become negligible as \( h \) approaches zero:

• \( \lim_{h \to 0} (6x^2h + 4xh^2 + h^3) = 0 \)

Thus, the expression simplifies to just \( 4x^3 \). The limit process removes the influence of the vanishing small parts, yielding the derivative, \( f'(x) = 4x^3 \). This powerful concept is essential because it lets us derive the exact rate of change of a function at any point \( x \).