Problem 26 - Calculus for Biology and Medicine

Question

Use the formal definition to find the derivative of $$ f(x)=\frac{1}{x+1} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Formal Definition of a Derivative

The formal definition of a derivative of a function \( f(x) \) at a point \( x = a \) is given by \( f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h} \). In this case, our function is \( f(x) = \frac{1}{x+1} \) and we want to find \( f'(x) \).

Step 2: Substitute the Function into the Derivative Formula

Substitute \( f(x) = \frac{1}{x+1} \) into the derivative formula to find \( f'(x) \). Thus, \( f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} = \lim_{{h \to 0}} \frac{{\frac{1}{x+h+1} - \frac{1}{x+1}}}{h} \).

Step 3: Simplify the Expression

To simplify, find a common denominator for \( \frac{1}{x+h+1} - \frac{1}{x+1} \). The common denominator is \( (x+h+1)(x+1) \). Thus, \( \frac{1}{x+h+1} - \frac{1}{x+1} = \frac{x+1 - (x+h+1)}{(x+h+1)(x+1)} = \frac{-h}{(x+h+1)(x+1)} \).

Step 4: Simplify the Derivative Expression Further

Substitute the simplified difference into the limit expression: \( f'(x) = \lim_{{h \to 0}} \frac{-h}{h (x+h+1)(x+1)} \). Simplify the fraction by canceling \( h \) from the numerator and the denominator, which results in \( f'(x) = \lim_{{h \to 0}} \frac{-1}{(x+h+1)(x+1)} \).

Step 5: Evaluate the Limit

Now, evaluate the limit as \( h \) approaches zero: \( f'(x) = \lim_{{h \to 0}} \frac{-1}{(x+h+1)(x+1)} = \frac{-1}{(x+1)^2} \). Thus, the derivative is \( f'(x) = \frac{-1}{(x+1)^2} \).

Key Concepts

Formal Definition of a Derivative

The formal definition of a derivative is a fundamental concept in calculus that helps us understand how a function changes at any given point. It is expressed as:

• \( f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h} \)

This definition essentially measures the rate of change of the function \( f(x) \) as \( x \) changes infinitesimally. By applying this definition, we can determine how the function changes at very small scales. For the function \( f(x) = \frac{1}{x+1} \), we use this formula to find its derivative, \( f'(x) \), which tells us how \( f(x) \) behaves as \( x \) shifts slightly.

Limit Process

When calculating a derivative using the formal definition, the limit process is essential. The idea is to consider what happens as the difference between two points on the function, \( h \), becomes very small, approaching zero. In our context, this means calculating:

• \( \lim_{{h \to 0}} \frac{{\frac{1}{x+h+1} - \frac{1}{x+1}}}{h} \)

Here, \( h \) is the minute change in \( x \), and we are interested in how the function \( f(x) \) changes with respect to this tiny increment. Using limits allows us to mathematically handle these tiny changes and find the derivative. In the case of \( f(x) = \frac{1}{x+1} \), the limit process will help us find the exact rate of change at any x value.

Simplifying Expressions

Simplifying mathematical expressions is crucial in calculations involving the derivative, as it makes the problem easier to solve. For our given function \( f(x) = \frac{1}{x+1} \), we initially have the expression:

• \( \frac{{\frac{1}{x+h+1} - \frac{1}{x+1}}}{h} \)

To simplify this, we find a common denominator for \( \frac{1}{x+h+1} \) and \( \frac{1}{x+1} \), which would be \((x+h+1)(x+1)\). This allows us to combine the fractions easily, leading to:

• \( \frac{-h}{(x+h+1)(x+1)} \)

Afterward, cancel out the \( h \) in the numerator and denominator (since \( h eq 0 \), as \( h \approx 0 \)), simplifying your expression to make it easier to evaluate the final limit. Thus, mastering this simplification step is key to solving derivatives accurately and efficiently.