Problem 104 - Thomas Calculus

Question

Suppose that \(u=g(x)\) is differentiable at \(x=-5, y=f(u)\) is differentiable at \(u=g(-5),\) and \((f \circ g)^{\prime}(-5)\) is negative. What, if anything, can be said about the values of \(g^{\prime}(-5)\) and \(f^{\prime}(g(-5)) ?\).

Step-by-Step Solution

Answer

One of \(g^{\prime}(-5)\) or \(f^{\prime}(g(-5))\) is negative, the other positive.

Step 1: Understanding the Composite Derivative

The derivative of a composite function at a point can be calculated using the Chain Rule. For two functions \(f(u)\) and \(g(x)\), their composite \((f \circ g)(x) = f(g(x))\), the derivative at any point is given by \((f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Here we apply this at \(x = -5\).

Step 2: Setting Up Equation Using Given Information

We know that \((f \circ g)^{\prime}(-5) = f^{\prime}(g(-5)) \cdot g^{\prime}(-5)\). It's given that \((f \circ g)^{\prime}(-5) < 0\). Therefore, we have the inequality \(f^{\prime}(g(-5)) \cdot g^{\prime}(-5) < 0\).

Step 3: Analyzing the Sign of the Product

For a product of two numbers \(a\) and \(b\) to be negative, one of them must be negative and the other must be positive. So, either \(f^{\prime}(g(-5)) < 0\) and \(g^{\prime}(-5) > 0\), or \(f^{\prime}(g(-5)) > 0\) and \(g^{\prime}(-5) < 0\).

Step 4: Conclusion on the Values

Based on the analysis, we conclude that one of the derivatives, \(g^{\prime}(-5)\) or \(f^{\prime}(g(-5))\), must be negative while the other must be positive.

Key Concepts

Composite Functions

A composite function is a combination of two functions where the output of one function becomes the input for another. Imagine we have two functions, say, \( g(x) \) and \( f(u) \). When we combine them, we create a new function denoted as \((f \circ g)(x) = f(g(x))\). This basically means you first apply \( g \), then use its output to apply \( f \).
Understanding composite functions is crucial for finding derivatives in calculus because we frequently encounter complex functions built this way. With composite functions, you often need to differentiate using a specialized technique known as the Chain Rule. This rule helps us handle the layering of operations in composite functions effectively. By breaking down each step, we can systematically derive a single derivative despite the complexity of the multiple layers of functions involved.

Derivative

The derivative of a function represents how the function's output changes as its input changes. It's like zooming in on a curve to see how steep or flat it really is at every point. Specifically, for a function \( f(x) \), the derivative \( f'(x) \) gives you the slope of the function at any point \( x \).
When it comes to differentiating composite functions, the Chain Rule is our friend. For example, when dealing with \((f \circ g)(x)\), the Chain Rule tells us that the derivative is \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \). This result is like saying, "First, see how much \( f(u) \) changes with respect to \( u \), and then multiply by how much \( g(x) \) changes with respect to \( x \)."

• In simpler terms, we're scaling the rate of change of \( f \) by the rate of change of \( g \).
• If \((f \circ g)'(-5)\) is negative, as in the exercise, it tells us that the overall rate of change of \((f \circ g)\) at \( x = -5 \) is in a downward direction. This insight is crucial in many applications, such as determining maximums, minimums, and the overall behavior of functions.

Differentiable Functions

A function is differentiable at a point if it has a derivative there, which implies it is smooth and has no sharp corners at that point. More formally, for a function \( f(x) \) to be differentiable at a point \( x = a \), it must be possible to define \( f'(a) \) such that \( f(x) \) smoothly transitions through \( a \).
In the given exercise, both \( u = g(x) \) and \( y = f(u) \) are differentiable at their respective points. This means both functions smoothly pass through \( x = -5 \) and \( u = g(-5) \), respectively. Differentiability is an important property because it assures continuity and smooth transitions of the function around that point.

• The Chain Rule relies heavily on the differentiability of both functions involved. Because \( f(u) \) and \( g(x) \) are differentiable, it guarantees that the composite function \((f \circ g)(x)\) is also differentiable.
• This property allows us to confidently apply the Chain Rule and understand the relationship between \( f'(g(-5)) \) and \( g'(-5) \). Knowing these functions are differentiable tells us the math will follow the expected continuous patterns.