Problem 14 - 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 \(x\). $$ S(x)=\frac{1}{x+1} $$

Step-by-Step Solution

Answer

The derivative is \( S'(x) = \frac{-1}{(x+1)^2} \).

Step 1: Define the Function and General Derivative Formula

First, we need to identify the given function and the formula to find the derivative. Here, the function is given as \( S(x) = \frac{1}{x+1} \). The derivative formula we will use is \( S'(x) = \lim_{h \to 0} \frac{S(x+h) - S(x)}{h} \).

Step 2: Find S(x+h)

Next, find the expression for \( S(x+h) \). Replace \( x \) with \( x+h \) in the function: \( S(x+h) = \frac{1}{(x+h)+1} = \frac{1}{x+h+1} \).

Step 3: Substitute into the Derivative Formula

Substitute \( S(x+h) \) and \( S(x) \) into the derivative formula:\[S'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h}\]

Step 4: Simplify the Expression Inside the Limit

Combine the two fractions in the numerator by getting a common denominator, \((x+h+1)(x+1)\):\[S'(x) = \lim_{h \to 0} \frac{(x+1) - (x+h+1)}{h(x+h+1)(x+1)}\]This simplifies to:\[S'(x) = \lim_{h \to 0} \frac{-h}{h(x+h+1)(x+1)}\]

Step 5: Cancel Common Factors

Notice the \( h \) in the numerator can cancel with one \( h \) in the denominator:\[S'(x) = \lim_{h \to 0} \frac{-1}{(x+h+1)(x+1)}\]

Step 6: Evaluate the Limit

As \( h \) approaches 0, the expression simplifies to:\[S'(x) = \frac{-1}{(x+1)(x+1)} = \frac{-1}{(x+1)^2}\]

Step 7: State the Derivative

Thus, the derivative of \( S(x) = \frac{1}{x+1} \) using the limit definition is:\[S'(x) = \frac{-1}{(x+1)^2}\]

Key Concepts

Limit Definition of Derivative

The foundation of understanding derivatives in calculus is the limit definition. This approach lets us discover the instantaneous rate of change of a function at any point. The derivative, according to this definition, is represented as follows:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]Here, \( h \) represents an infinitely small increment in \( x \), allowing us to calculate the slope of the tangent line to the function at point \( x \). By evaluating this limit, we're essentially checking how the function value changes as \( h \) gets closer and closer to zero.### Why Use the Limit Definition?- **Understanding Change:** This definition gives a precise way of calculating how a function changes at any given point.- **Foundational Concepts:** It's essential for diving deeper into calculus and helps establish fundamental calculus ideas.The exercise illustrates this concept by finding the derivative of the function \( S(x) = \frac{1}{x+1} \) using the limit definition, which reveals how the function behaves locally at any value of \( x \). Applying this definition, you get to see the nuts and bolts of how derivatives work, especially in the context of rational functions like this one.

Rational Functions Derivative

Rational functions are quotients of polynomials, like \( \frac{1}{x+1} \) in the exercise provided. Finding the derivative of a rational function using the limit definition involves a sequence of steps commonly needed to simplify expressions and evaluate the limit.### Key Steps for Derivative Calculation- **Substitute \( x + h \):** Insert \( x + h \) in place of \( x \) in the original function to get the new expression, \( S(x+h) = \frac{1}{x+h+1} \).- **Simplify the Difference Quotient:** The expression \( \frac{S(x+h) - S(x)}{h} \) often leads to complex fractions. We simplify this to enable efficient limit evaluation, often by finding a common denominator.- **Cancel Factors:** After simplification, canceling out terms, especially \( h \), helps reach the final simplified form.When dealing with rational functions, it's crucial to watch signs and simplify carefully to avoid mistakes, as shown when we arrived at the derivative \( S'(x) = \frac{-1}{(x+1)^2} \). Rational functions typically involve algebraic manipulations that highlight their unique challenges and intricacies.

First Principles Derivative

The idea behind the first principles derivative is to derive the derivative from primary mathematical principles, without shortcuts or pre-established rules. In essence, it's another name for the limit definition of the derivative. By deriving from first principles, you gain a deeper and more foundational understanding of calculus.### First Principles in ActionBy applying the limit \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) directly, the exercise demonstrates the step-by-step simplification necessary to solve for the derivative. This involves:- **Substituting into Formula:** Perform substitutions with care to ensure each part of the function comes together systematically.- **Systematic Simplification:** Using algebra to simplify complex fractions or polynomials ensures the expression can be evaluated.The simplicity of the first principles approach is in how basic algebra interacts with limits. It's the bedrock of understanding how derivatives function mathematically. In our example, this method led to the derivative \( \frac{-1}{(x+1)^2} \), showcasing how even straightforward rational functions can be tackled from core mathematical principles.