Problem 69 - Calculus Single Variable

Question

Assuming that the range of \(f\) is contained in the domain of \(g,\) find a formula for \((g \circ f)^{\prime \prime}(c) .\) Suppose that \(f\) and \(g\) are concave up. What property of \(g\) ensures that \(g \circ f\) is also concave up?

Step-by-Step Solution

Answer

The condition \(g' \geq 0\) ensures the composite function \(g \circ f\) remains concave up.

Step 1: Understanding Composite Function

When finding the derivative of a composite function \((g \circ f)(x) = g(f(x))\), we must apply the chain rule. First, identify the need to use the chain rule twice for the second derivative.

Step 2: Finding the First Derivative

Apply the chain rule to find the first derivative of the composite function. The first derivative is given by:\((g \circ f)'(x) = g'(f(x)) \cdot f'(x)\).

Step 3: Finding the Second Derivative

To find the second derivative, differentiate \(g'(f(x)) \cdot f'(x)\) again with respect to \(x\). Use the product rule and chain rule:\[(g \circ f)''(x) = g''(f(x)) \cdot (f'(x))^2 + g'(f(x)) \cdot f''(x)\].

Step 4: Analyzing Concavity of Functions

Given that \(f\) and \(g\) are concave up, this means their second derivatives are non-negative: \(f''(x) \geq 0\) and \(g''(y) \geq 0\) for all relevant \(x\) and \(y\).

Step 5: Ensuring Concavity of Composite Function

For \(g \circ f\) to be concave up, its second derivative should be non-negative:\[(g \circ f)''(x) = g''(f(x)) \cdot (f'(x))^2 + g'(f(x)) \cdot f''(x) \geq 0\].Since \(f'(x)^2 \geq 0\) and \(f''(x) \geq 0\), having \(g'(y) \geq 0\) ensures the whole expression stays non-negative.

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus. It helps us differentiate composite functions, such as \((g \circ f)(x) = g(f(x))\). When you have a function inside another function, think of the chain rule as a method to "unravel" the layers of these functions to find the derivative.

Here's how it works: if you want the derivative of \(g(f(x))\), you first differentiate the outer function \(g\) with respect to the inner function \(f(x)\), and then multiply by the derivative of the inner function \(f(x)\). That's how you get \(g'(f(x)) \cdot f'(x)\).

In the context of our problem, finding the second derivative involves applying the chain rule twice. It requires careful attention to both layers of the composite function. Essentially, once you've differentiated the first time, you treat the results as separate functions and then apply the chain rule again. This step is crucial for accurately unraveling and differentiating composite functions.

Product Rule

When working with second derivatives of composite functions, the product rule often comes into play. The product rule is essential when you have two functions multiplied together and you need to find their derivative.

In simple terms, if you have functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x) \cdot v(x)\) is given by: \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\). You differentiate each function separately and then apply this formula.

In our exercise, during the second derivative calculation, we use the product rule on \(g'(f(x)) \cdot f'(x)\). This calculates the derivative of this product, and then follows up by applying the chain rule again.

Understanding and combining the chain and product rules helps navigate through complex expressions when differentiating composite functions.

Concavity

Concavity is an important concept in calculus that describes the direction of a curve. If a function is concave up, like a cup, it means the function's slope increases as you move along the graph. Conversely, a concave down function looks like a frown.

Mathematically, a function is concave up if its second derivative is non-negative. This is because the second derivative measures the rate of change of the slope or how the slope itself changes. If \(f''(x) \geq 0\), the curve is upwards bending.

In our scenario, both \(f\) and \(g\) are given as concave up functions. This information helps in predicting the behavior of \(g \circ f\). To ensure \(g \circ f\) is also concave up, its second derivative must be non-negative, reaffirmed by the individual second derivatives of \(f\) and \(g\). This reflects their inherent upward-bending nature, contributing to the overall concavity of the composite function.

Second Derivative

The second derivative provides us with insights about the acceleration of a function's slope. It tells us whether a function is curving up or down, much like the concavity.

In calculus, the second derivative of a composite function can get complicated, as it involves multiple applications of differentiation concepts. For our problem, the second derivative of \(g(f(x))\) involves differentiating the result of the first derivative: \(g'(f(x)) \cdot f'(x)\).

Calculating the second derivative here means using both the chain rule and the product rule together:

• First, apply the product rule to \(g'(f(x)) \cdot f'(x)\).
• Then, within this process, apply the chain rule again to each component as needed.

This combined approach gives us \((g \circ f)''(x) = g''(f(x)) \cdot (f'(x))^2 + g'(f(x)) \cdot f''(x)\). Each piece of this expression provides crucial information, particularly in contexts like assessing concavity.