Problem 65 - Calculus and its applications

Question

Let \(f(u)=\sqrt[3]{u}\) and \(g(x)=u=1+3 x^{2}\). Find \((f \circ g)^{\prime}(2)\).

Step-by-Step Solution

Answer

\((f \circ g)'(2) = 4 \cdot (13)^{-2/3}\).

Step 1: Understand the Function Composition

The composition \((f \circ g)(x)\) represents \(f(g(x))\). Given \(f(u) = \sqrt[3]{u}\) and \(g(x) = 1 + 3x^2\), we can write \((f \circ g)(x) = \sqrt[3]{1 + 3x^2}\).

Step 2: Differentiate the Outer Function

First, let's find the derivative of \(f(u) = u^{1/3}\). Using the power rule, \(f'(u) = \frac{1}{3}u^{-2/3}\).

Step 3: Differentiate the Inner Function

Now we need the derivative of \(g(x) = 1 + 3x^2\). Differentiating term by term, we get \(g'(x) = 6x\).

Step 4: Apply the Chain Rule

The chain rule tells us \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\). Substitute the derivatives from Steps 2 and 3: \[(f \circ g)'(x) = \left(\frac{1}{3}(1 + 3x^2)^{-2/3}\right) \cdot (6x)\].

Step 5: Evaluate the Derivative at x=2

Substitute \(x = 2\) into the derived formula:\((f \circ g)'(2) = \frac{1}{3}(1 + 3(2)^2)^{-2/3} \cdot (6 \cdot 2)\).Calculate:\((f \circ g)'(2) = \frac{1}{3}(13)^{-2/3} \cdot 12\).

Step 6: Simplify

Further simplify:\((f \circ g)'(2) = 4 \cdot (13)^{-2/3}\).This is the final simplified expression.

Key Concepts

Function Composition

Function composition in calculus is like nesting two functions within each other. It helps us understand how one function affects the behavior of another function. Imagine having two functions, say \( f(u) \) and \( g(x) \). The notation \((f \circ g)(x)\) means you're applying \( g(x) \) first, and then using its output as an input for \( f(u) \). In the given example, \((f \circ g)(x) = \sqrt[3]{1 + 3x^2}\), you're essentially calculating \( g(x) = 1 + 3x^2 \) first and then taking this result into the cube root function \( f(u) = \sqrt[3]{u}\). By considering this setup, you can analyze how changes in \( x \) impact the final outcome of the composed function.

Power Rule

The power rule in calculus is a shortcut that tells us how to differentiate functions of the form \( u^n \). This rule states that if \( f(u) = u^n \), then the derivative \( f'(u) = n \cdot u^{n-1} \). It greatly simplifies taking derivatives, especially when dealing with polynomial expressions.Consider the function \( f(u) = \sqrt[3]{u} \), which can be rewritten as \( u^{1/3} \). By applying the power rule, the derivative becomes \( f'(u) = \frac{1}{3} \cdot u^{-2/3} \). This transformation makes it straightforward to handle calculus operations, easing the complexity particularly when such functions are embedded in larger expressions or compositions.

Differentiation

Differentiation is the process of finding the derivative of a function, which gives us the rate at which the function's output changes with respect to its input. It's a fundamental operation in calculus used to analyze the behavior of mathematical models.To differentiate a function like \( g(x) = 1 + 3x^2 \), you break it down into smaller parts. In this example, by differentiating term by term, you calculate that the derivative \( g'(x) = 0 + 6x \), leveraging the power of simple rules like differentiating constants to zero and power derivatives. This result tells us how the function \( g(x) \) behaves at any point \( x \), delivering insights into how small changes in \( x \) will impact the value of \( g(x) \).Differentiation not only decodes single functions, but also serves as a crucial tool for advanced concepts like the chain rule, where multiple layers of functions interweave. Understanding how to differentiate enables you to bypass intricate computations and focus more on the purpose and application of calculus.