Problem 249 - Calculus Volume 1

Question

Use the information in the following table to find \(h^{\prime}(a)\) at the given value for \(a\). $$\begin{array}{|c|c|c|c|c|} \hline x & f(x) & f^{\prime}(x) & g(x) & g^{\prime}(x) \\ \hline 0 & 2 & 5 & 0 & 2 \\ \hline 1 & 1 & -2 & 3 & 0 \\ \hline 2 & 4 & 4 & 1 & -1 \\ \hline 3 & 3 & -3 & 2 & 3 \\ \hline \end{array}$$ $$ h(x)=f(x+f(x)) ; a=1 $$

Step-by-Step Solution

Answer

The derivative \(h'(1)\) is \(-4\).

Step 1: Substitute into Inner Function

We need to calculate the value of the inner function of \(h(x)\), which is \(x + f(x)\). For \(x = 1\), find \(f(1)\) from the table, which is 1. Thus, substitute into the inner function: \(1 + f(1) = 1 + 1 = 2\).

Step 2: Differentiate the Outer Function

To find the derivative \(h'(x)\), we apply the chain rule. The derivative \(h'(x) = f'(x + f(x)) \cdot (1 + f'(x))\). Here, we substitute \(x = 1\) to get \(1 + f'(1) = 1 + (-2) = -1\). Now, for the outer function derivative: \(f'(x + f(x)) = f'(2)\), which is given as 4 in the table.

Step 3: Calculate the Derivative at Given Point

Now we'll compute \(h'(1) = f'(2) \cdot (-1)\). Based on the values we found: \(f'(2) = 4\) and the factor from the chain rule is \(-1\). Hence, \(h'(1) = 4 \times (-1) = -4\).

Key Concepts

chain rule

In calculus, the chain rule is an essential tool for differentiating composite functions, such as those with nested or multiple layers. The rule allows us to find the derivative of composite functions by addressing each layer individually.

When you have a function like \( h(x) = f(g(x)) \), the chain rule states that the derivative \( h'(x) \) is given by \( f'(g(x)) \cdot g'(x) \). In simple terms, you first take the derivative of the outer function \( f \) with respect to its inner function \( g(x) \), and then multiply it by the derivative of the inner function \( g \) itself.

For the given exercise, the focus is on differentiating the function \( h(x) = f(x + f(x)) \). Using the chain rule here, we treat \( x + f(x) \) - the inner function - separately. The derivative of this inner part is \( 1 + f'(x) \). We then multiply by the derivative of the outer function, which has its derivative evaluated at the result of the inner function, which in this case is \( f'(x + f(x)) \).

function tables

A function table is a structured way of presenting values for a function along with its derivatives. It provides clear insights into how a function behaves at specific points.

For example, the table from the exercise presents point values for \( x \), \( f(x) \), and derivatives \( f'(x) \) and \( g'(x) \). These values are crucial in computing derivatives and applying the chain rule effectively.

Using a function table, you can:

• Identify values of the function and its derivative at specific points
• Find out how fast the function is changing at a given point, which is essential for calculating derivatives
• Support analyses of complex functions by breaking them down into simpler forms

computing derivatives

Computing derivatives is a core aspect of calculus that involves finding the rate at which a function changes at a specific point. Understanding how to compute derivatives is vital for analyzing the dynamics of mathematical models and systems.

In this exercise, calculating \( h'(a) \) for \( h(x) = f(x + f(x)) \) involves both differentiation and substitution.

First, determine the expression of the composite function derivative using the chain rule. Then, substitute the given point to compute \( h'(a) \). This process illustrates that computing derivatives isn't just analytical but also involves evaluating specific values from tables or given data.

Remember:

• Derivatives represent the slope or rate of change, critical for understanding function behavior
• Substitute the values given for \( x \) and \( a \) properly to reconstruct the function's behavior at those points
• Each computation step relies on understanding both the function and the context provided