Problem 44 - Calculus Single Variable

Question

Suppose that \(f, g,\) and \(h\) are differentiable functions and that \(f=g \cdot h\). Suppose that \(g(4)=6, d g(4)=7, d x, h(4)=\) \(5,\) and \(d h(4)=-2 d x .\) What is \(d f(4) ?\)

Step-by-Step Solution

Answer

The differential \(df(4)\) is \(23\, dx\) or numerically, 23.

Step 1: Identify the Given Information

We are given that the functions \(f\), \(g\), and \(h\) are differentiable, and \(f = g \cdot h\). Also, we have specific values at \(x = 4\): \(g(4) = 6\), \(dg(4) = 7\, dx\), \(h(4) = 5\), and \(dh(4) = -2\, dx\).

Step 2: Understand the Problem

To find \(df(4)\), the differential at \(x = 4\), we need to use the product rule because \(f(x) = g(x) h(x)\). The derivative of a product \(f = g \cdot h\) is given by \(df = g \cdot dh + h \cdot dg\).

Step 3: Apply the Product Rule

Using the product rule \(df = g \cdot dh + h \cdot dg\), we substitute the given values: \(g = 6\), \(dg = 7\, dx\), \(h = 5\), and \(dh = -2\, dx\). This gives us: \[ df = 6(-2\, dx) + 5(7\, dx) \]

Step 4: Simplify the Expression for df(4)

Substitute \(dx\) as a common factor: \[ df = 6(-2)\, dx + 5(7)\, dx = (-12 + 35)\, dx = 23\, dx \] So, \(df(4) = 23\, dx\).

Step 5: Write the Final Answer

The change in \(f\) at \(x = 4\) or \(df(4)\) is \(23\, dx\), which equals \(23\) as \(dx\) is the small change in \(x\), often treated as 1 in such calculations.

Key Concepts

Product Rule

In calculus, the product rule is an essential technique for finding the derivative of a product of two differentiable functions. When you have a function like \( f(x) = g(x) \cdot h(x) \), the product rule allows you to differentiate it effectively. Instead of applying the basic rules for differentiation to a product directly, the product rule provides a formula:\[\frac{d}{dx}(g \cdot h) = g \cdot \frac{dh}{dx} + h \cdot \frac{dg}{dx}\]This means the derivative of the product \( g(x) \cdot h(x) \) is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. This approach simplifies the process and ensures accurate results. Always remember:

• You need both functions to be differentiable for the product rule to apply.
• Carefully substitute each part with the appropriate derivatives and function values.
• The product rule is essential for problems where functions are multiplied together, such as in physics or engineering problems.

Differentiable Functions

Differentiable functions are a key concept in calculus. Simply put, a function is differentiable at a point if it has a derivative there. This means the slope of the tangent line to the graph of the function is well-defined at that point. For differentiation to occur, a few conditions need to be satisfied:

• The function must be continuous at the point concerned—there should be no breaks, jumps, or holes in the graph.
• The limit of the function's derivative should exist as you approach the point from either side.

Differentiability is important because it ensures that you can apply various calculus techniques, like the product rule, to solve practical problems involving rates of change and slopes of curves. In our exercise, understanding that \( f \), \( g \), and \( h \) are differentiable allows us to confidently apply the product rule.While continuity and differentiability often go hand in hand, it is critical to note that existence of one does not automatically imply the other, except in certain contexts.

Calculus Problem Solving

Approaching calculus problems, especially those involving differentiation, requires a structured method to simplify and effectively solve them. The first step is understanding the problem statement. This involves identifying known values and what you need to find out. Ensure all the functions involved are differentiable at the points of interest. Calculus problem solving often involves these strategies: - **Break down the problem:** Understand which differentiation rules apply. For a product of functions, this means applying the product rule. - **Substitution:** Use given values and derivatives to perform substitutions as needed, making the computation straightforward. - **Simplification:** Carefully simplify the expressions to arrive at the simplest form of the derivative. Using these methods allows you to tackle calculus equations systematically, as was done in the exercise. Applying these steps justly reduces complex calculus problems into manageable calculations, equipping you to handle various scenarios in science and engineering endeavors.