Problem 60 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Suppose \(h=f \circ g .\) Show that \(h^{\prime}=\left(f^{\prime} \circ g\right) g^{\prime}\).

Step-by-Step Solution

Answer

The derivative of the composite function \(h = f \circ g\) is given by the Chain rule as \(h'(x) = f'(g(x)) \cdot g'(x)\). Rewriting this in the product form, we have \[h^{\prime} = \left(f^{\prime} \circ g\right)g^{\prime}.\]

Step 1: Write the given composite function

We are given that the composite function is \(h = f \circ g\). This means that \[h(x) = f(g(x)).\]

Step 2: Apply the Chain rule

Now, we need to find the derivative of \(h\) with respect to its input variable \(x\). Using the Chain rule, we have \[h'(x) = f'(g(x)) \cdot g'(x).\] Here, \(f'(g(x))\) represents the derivative of the outer function, \(f\), evaluated at the inner function, \(g(x)\), and \(g'(x)\) represents the derivative of the inner function.

Step 3: Write out the result

We can now rewrite the result as \[h'(x) = \left(f^{\prime} \circ g\right)g^{\prime}(x),\] where \(\left(f^{\prime} \circ g\right)g^{\prime}\) is the product of \(\left(f^{\prime} \circ g\right)\) and \(g^{\prime}\). This proves that \[h^{\prime} = \left(f^{\prime} \circ g\right)g^{\prime}.\]

Key Concepts

Composite Functions

Composite functions are an essential concept in calculus, often represented as a combination of two or more functions. To explain simply, when you have a function like \(g(x)\) and another function \(f(x)\), a composite function \(h(x)\) is created by composing them, generally written as \(h = f \circ g\) or \(h(x) = f(g(x))\). This means you first apply \(g\) to \(x\), and then \(f\) to the result of \(g(x)\). A good way to understand this concept is to think of it as a two-step process where you're plugging one function into another. The composition can create more complex relationships, showing how one quantity depends on another through a series of steps.

• This technique helps in simplifying expressions and solving equations.
• Understanding how to combine functions leads to insights in modeling real-world problems.

As an example, if \(g(x) = x^2\) and \(f(y) = y + 3\), the composite function would be \(f(g(x)) = (x^2) + 3\). Here, you can see how the composition \(f \circ g\) affects the input \(x\). By grasping how composite functions operate, we unlock the ability to tackle more complicated scenarios in calculus.

Derivatives

Derivatives are one of the fundamental building blocks of calculus. In simple terms, a derivative represents how much a function changes as its input changes, essentially providing the rate at which things happen. It gives us a way to understand the slope of a tangent line to a curve at any point. Think of it as a mathematical tool used to calculate how fast something is increasing or decreasing.
If you consider a function \(y = f(x)\), the derivative is represented by \(f'(x)\) or \(\frac{dy}{dx}\). It tells us the rate of change of \(y\) with respect to \(x\).

• Derivatives help in understanding increase and decrease within a function.
• They are used to find maxima and minima values of curves, which is crucial in optimizing processes.

In the context of composite functions, the chain rule—a crucial rule for finding derivatives—comes into play. Here, if \(h(x) = f(g(x))\), then the derivative \(h'(x)\) involves differentiating both \(f\) and \(g\), showing how changes in \(g\) influence changes in \(f\). This connection helps in solving complex differentiation problems in calculus.

Mathematical Proofs

Mathematical proofs are rigorous arguments that verify the truth of mathematical statements. In calculus, proofs help confirm the rules and theories we rely upon to solve problems, such as the chain rule we discussed above. A proof provides a logical series of steps showing why a particular theorem is true.
In the original problem, the chain rule is applied to prove the relationship \(h^{\prime} = \left(f^{\prime} \circ g\right)g^{\prime}\). The process involves:

• Identifying the composite function as \(h(x) = f(g(x))\).
• Using the chain rule to differentiate \(h(x)\) into \(f'(g(x)) \cdot g'(x)\).
• Rewriting the expression as \(\left(f^{\prime} \circ g\right)g^{\prime}(x)\), completing the proof.

Proofs are important because they ensure that the mathematics we use is based on solid foundational principles. They help in understanding not just the 'how' but also the 'why' behind mathematical operations. By developing proof skills, students can enhance their logical thinking and deepen their understanding of calculus concepts.