Problem 55 - Calculus an Applied Approach

Question

determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f^{\prime}(c)\) and \(g^{\prime}(c)\) are zero and \(h(x)=f(x) g(x),\) then \(h^{\prime}(c)=0\)

Step-by-Step Solution

Answer

The statement is true. When both \(f^{\prime}(c)\) and \(g^{\prime}(c)\) are zero, the expression for \(h^{\prime}(c)\) simplifies to zero, regardless of the values of \(f(c)\) and \(g(c)\).

Step 1: Applying The Product Rule to \(h(x)\)

According to the product rule of derivatives, the derivative of the function \(h(x) = f(x)g(x)\) is given by \[h^{\prime}(x) = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)\]

Step 2: Determine \(h^{\prime}(c)\)

Substituting \(x = c\) in the above expression, we get \[h^{\prime}(c)= f^{\prime}(c)g(c) + f(c)g^{\prime}(c)\]

Step 3: Analyze givens

We are given that \(f^{\prime}(c)\) and \(g^{\prime}(c)\) are both zero. When we substitute these values into the equation for \(h^{\prime}(c)\), it simplifies as \[h^{\prime}(c)= 0*g(c) + f(c)*0 = 0\].

Key Concepts

Derivative

The concept of a derivative is a cornerstone in calculus, representing the rate at which a function is changing at any given point. Specifically, the derivative of a function at a point measures the instantaneous rate of change, akin to finding the velocity of an object at a specific moment in time. For a function represented as \(y = f(x)\), the derivative at \(x = c\), denoted as \(f^{\text{'}}(c)\), is the limit of the average rate of change as the interval approaches zero.

Understanding derivatives is fundamental to various branches of mathematics and physics since they describe dynamic changes in systems. Whether dealing with acceleration in physics, slope of a tangent in geometry, or even marginal cost in economics, the derivative is an essential tool for analysis and problem-solving.

Calculus

Calculus is a vast field of mathematics that deals with continuous change. It's divided into two main branches: differential calculus and integral calculus. Differential calculus, which encompasses the concept of derivatives, focuses on dividing things into small (infinitesimally small) pieces to analyze how they change, while integral calculus pieces these infinitesimal parts together to find totals such as areas, volumes, and accumulations. The two branches are connected by the Fundamental Theorem of Calculus, establishing an intricate relationship between differentiation and integration.

The use of calculus extends beyond academic pursuits, deeply ingrained in engineering, physics, economics, and even in emerging fields like machine learning and data science. Its applications are vast, turning it into a critical tool for modeling and solving real-world scenarios where change is a constant factor.

Product Rule

When faced with the task of differentiating the product of two functions, simply taking the derivative of each individually and multiplying them will not yield the correct result. This is where the product rule comes in. It states that for two differentiable functions, \(f(x)\) and \(g(x)\), the derivative of their product \(h(x) = f(x)g(x)\) is given by
\[h^{\text{'}}(x) = f^{\text{'}}(x)g(x) + f(x)g^{\text{'}}(x)\].

This rule is essential when dealing with products in differentiation because it accurately accounts for the interaction between the two functions as they change. For example, when a physical problem involves forces that change with respect to time and distance simultaneously, using the product rule can accurately determine the rate of work done over time.

Differentiation

Differentiation is the process of finding a derivative. It involves taking a function and finding the formula that provides the rate of change of that function. Differentiation can be performed using various rules and techniques, including the product rule, the quotient rule, the chain rule, and more.

Through differentiation, we can solve a wide range of problems, like finding maximum and minimum values of functions (important in optimization problems), analyzing the motion of objects (in kinematics), and determining rates of reaction in chemistry. Proper understanding and application of differentiation are crucial for solving problems involving rates of change in nearly any field of science and engineering.

• The process is systematic and often follows a set of known rules.
• Each rule has its application based on the function's composition.
• Understanding when and how to apply each rule is key to mastering differentiation.

As you engage more deeply with calculus, these principles become intuitive, and the power of differentiation in modeling and problem-solving becomes even more apparent.