Problem 29 - Calculus Single Variable

Question

Compute \((f \circ g)^{\prime}\) and \((g \circ f)^{\prime}\). $$ f(x)=x /(x+1), g(x)=x^{3} $$

Step-by-Step Solution

Answer

\((f \circ g)'(x) = \frac{3x^2}{(x^3 + 1)^2}, \ (g \circ f)'(x) = \frac{3x^2}{(x+1)^4}.\)

Step 1: Understand Composite Functions

The notation \((f \, \circ \, g)(x)\) represents the composition of functions \(f\) and \(g\), meaning \(f(g(x))\). Similarly, \((g \, \circ \, f)(x)\) means \(g(f(x))\). Our task is to find the derivatives of these composite functions.

Step 2: Find the Composition \((f \, \circ \, g)(x)\)

Substitute \(g(x) = x^3\) into \(f(x) = \frac{x}{x+1}\) to get \((f \, \circ \, g)(x) = f(x^3) = \frac{x^3}{x^3 + 1}\).

Step 3: Differentiate \((f \, \circ \, g)(x)\) using the Quotient Rule

The derivative of \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \). Here, \(u = x^3\) and \(v = x^3 + 1\). Thus, \( u' = 3x^2 \) and \( v' = 3x^2 \). Apply the quotient rule: \[(f \, \circ \, g)^{\prime}(x) = \frac{(3x^2)(x^3 + 1) - (x^3)(3x^2)}{(x^3 + 1)^2} = \frac{3x^5 + 3x^2 - 3x^5}{(x^3 + 1)^2} = \frac{3x^2}{(x^3 + 1)^2}.\]

Step 4: Find the Composition \((g \, \circ \, f)(x)\)

Substitute \(f(x) = \frac{x}{x+1}\) into \(g(x) = x^3\) to get \((g \, \circ \, f)(x) = g\left(\frac{x}{x+1}\right) = \left(\frac{x}{x+1}\right)^3.\)

Step 5: Differentiate \((g \, \circ \, f)(x)\) using the Chain Rule

To find the derivative, use the chain rule: if \(y = u^3\), then \(\frac{dy}{dx} = 3u^2 \frac{du}{dx}\), where \(u = \frac{x}{x+1}\). Here, \(\frac{du}{dx} = \frac{1}{(x+1)^2}\) using the quotient rule. So, \[(g \, \circ \, f)^{\prime}(x) = 3\left(\frac{x}{x+1}\right)^2 \cdot \frac{1}{(x+1)^2} = \frac{3x^2}{(x+1)^4}.\]

Key Concepts

Composite Functions

When it comes to calculus, a composite function is a very useful concept. It occurs when you have two functions and combine them into a single new function. Think of it as plugging the output of one function directly into the input of another. For example, let's take our two functions, \( f(x) \) and \( g(x) \). A composite function \( (f \circ g)(x) \) means that you first apply the function \( g(x) \), then take whatever result you get and put it through \( f(x) \). In our exercise, we have \( f(x) = \frac{x}{x+1} \) and \( g(x) = x^3 \). The composite \( (f \circ g)(x) \) becomes substituting \( g(x) \) into \( f(x) \), so it's like nesting one function inside another like this:

• Perform \( g(x) = x^3 \)
• Then, substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = \frac{x^3}{x^3 + 1} \)

Understanding how composite functions work lays the foundation for more complex calculus operations.

Quotient Rule

The quotient rule is like a trusty mathematical tool that helps us when differentiating functions that are written as one function divided by another. To put it simply, it helps us find the derivative of a division of two functions, \( \frac{u}{v} \).The rule says that the derivative of \( \frac{u}{v} \) is given by: \[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]In our specific case, we applied the quotient rule to find the derivative of \( (f \circ g)(x) \). Here, \( u = x^3 \) and \( v = x^3 + 1 \). Step-by-step, we go:

• Find the derivative \( u' = 3x^2 \)
• Then, find \( v' = 3x^2 \)
• Apply the rule: \( \frac{(3x^2)(x^3 + 1) - (x^3)(3x^2)}{(x^3 + 1)^2} = \frac{3x^2}{(x^3 + 1)^2} \)

This approach ensures you get the correct result, avoiding potential algebraic pitfalls along the way.

Chain Rule

The chain rule is used in calculus to differentiate composite functions. It's a method that allows us to take the derivative of a "chain" of functions linked together. Imagine a function inside another, creating a layered effect. The chain rule is essential for breaking down these layers and calculating derivatives efficiently. Conceptually, if \( y = u(v(x)) \), the chain rule tells us:\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx}\]In our exercise, we apply the chain rule to \((g \circ f)(x)\) where \( g(x) = x^3 \) and \( f(x) = \frac{x}{x+1} \).

• First, we substitute to get \( g\left(\frac{x}{x+1}\right) = \left(\frac{x}{x+1}\right)^3 \)
• Applying the chain rule involves differentiating the outer function, then multiplying by the derivative of the inner function.
• Here, \( (g \circ f)'(x) = 3\left(\frac{x}{x+1}\right)^2 \cdot \frac{1}{(x+1)^2} = \frac{3x^2}{(x+1)^4} \)

The chain rule simplifies dealing with composite structures, making it a cornerstone of differentiation techniques in calculus.