Problem 251 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The derivative \( h'(1) \) is \(-12\).

Step 1: Identify the Function to Differentiate

We are given the function \( h(x) = g(2 + f(x^2)) \) and we need to find \( h'(a) \) where \( a = 1 \). This involves both chain and nested function rules of differentiation.

Step 2: Differentiate Using the Chain Rule

Apply the chain rule to differentiate \( h(x) \). The outer function is \( g(u) \), where \( u = 2 + f(v) \) and \( v = x^2 \). The derivative is \( h'(x) = g'(u) \cdot (2 + f(v))' \).

Step 3: Differentiate the Inner Functions

First, differentiate \( 2 + f(v) \) with respect to \( v \): we get \( f'(v) \). Then differentiate \( v = x^2 \) with respect to \( x \): we get \( 2x \). Thus, \( (2 + f(x^2))' = f'(x^2) \cdot 2x \).

Step 4: Calculate \( h'(x) \) Using Given Values

Substitute \( x = 1 \) into \( h'(x) = g'(2 + f(x^2)) \cdot f'(x^2) \cdot 2x \). First, calculate \( x^2 = 1^2 = 1 \). At \( v = 1 \), from the table, \( f(v) = f(1) = 1 \), \( f'(v) = f'(1) = -2 \). Substitute these into \( 2 + f(1) = 2 + 1 = 3 \). Then at \( u = 3 \), \( g'(u) = g'(3) = 3 \).

Step 5: Compute the Final Derivative Value

Substitute the known values into the expression: \( h'(1) = g'(3) \cdot f'(1) \cdot 2 \cdot 1 \). This gives \( h'(1) = 3 \cdot (-2) \cdot 2 = -12 \).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential tool in calculus, primarily used when differentiating compositions of functions. It's like pulling apart a puzzle, step by step. Imagine you're peeling an onion; you start from the outermost layer, slowly working inward.To apply the chain rule, follow these basic steps:

• Identify the outer function and the inner function(s) within the composed function.
• Differentiate the outer function and multiply it by the derivative of the inner function.

In our exercise, we have the function: \[h(x) = g(2 + f(x^2))\]Here, we have layers upon layers: an inner function \(f(x^2)\) inside the function \(g\). The chain rule helps us differentiate by tackling each part one piece at a time.

Nested Functions Explained

Nested functions are layers of functions within each other, like a set of Russian dolls. In our example, the function \(h(x)\) involves the nesting of functions \(g\) and \(f\). To break it down, first look at the innermost function:

• \(x^2\) – This is the base, the simplest part.

Then, we have:

• \(f(x^2)\) – \(f\) is acting upon \(x^2\), wrapping it to form a slightly larger function.
• \(2 + f(x^2)\) – Here, \(2\) is added to \(f(x^2)\), creating another layer.
• Finally, \(g(2 + f(x^2))\) – This is the outer shell, where \(g\) acts on the combined result.

Each layer adds complexity, but by using the chain rule, we peel back these layers to simplify differentiation.

Derivative Calculation Process

Calculating the derivative of a function with nested functions uses systematic steps: start from the outer function and work your way inward.**Steps to Calculate Derivatives:**1. **Differentiate the Outer Function:** Start by differentiating the outermost function in combination, like \(g\) in our example.2. **Inner Differentiation:** Move inward, calculating the derivatives of inner functions such as \(2 + f(x^2)\). This often involves finding derivatives for expressions like \(f'(x^2)\) and using basic derivative rules for powers and sums.3. **Combine Derivatives:** Multiply the derivative of the outer function by the derivatives of each nested function.In our problem, once differentiated, everything was substituted using given values from the table.

Using a Table Lookup

A table lookup involves referencing values from a provided table to supplement calculus problems, which saves time and maintains accuracy.In this exercise:- The table gives values for functions \(f\), \(g\) and their derivatives \(f'\) and \(g'\) at specific points.- For instance, at \(x=1\), check values: \(f(1)\), \(f'(1)\), \(g(3)\), and \(g'(3)\) directly from the table.Using these pre-calculated values reduces errors and speeds up the derivative calculation process. It is especially helpful for nested functions where pinpointing exact values manually would be much more complex and time-consuming.