Problem 70 - Calculus and its applications

Question

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$ f(x)=\sqrt[3]{2 x+\left(x^{2}+x\right)^{4}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{1}{3}\left(2x + (x^2 + x)^4\right)^{-2/3}(2 + 4(x^2 + x)^3(2x + 1)) \).

Step 1: Identify Inner and Outer Functions

The function inside the cube root is the inner function, and the cube root itself is the outer function. We have the outer function as \( g(u) = u^{1/3} \) and the inner function as \( u(x) = 2x + (x^2 + x)^4 \).

Step 2: Differentiate the Outer Function

Differentiate \( g(u) = u^{1/3} \) with respect to \( u \). The derivative is \( g'(u) = \frac{1}{3}u^{-2/3} \).

Step 3: Differentiate the Inner Function

Now differentiate \( u(x) = 2x + (x^2 + x)^4 \) with respect to \( x \).We need to apply the chain rule again on \( (x^2 + x)^4 \).- Let \( v(x) = x^2 + x \), then \( (x^2 + x)^4 \) becomes \( v(x)^4 \).- Differentiate \( v(x)^4 \) to get \( 4v(x)^3v'(x) \), where \( v'(x) = 2x + 1 \).Therefore, \( (x^2 + x)^4 \) differentiates to \( 4(x^2 + x)^3(2x + 1) \).- Differentiate \( 2x \) to get \( 2 \).Now, combining these, \( u'(x) = 2 + 4(x^2 + x)^3(2x + 1) \).

Step 4: Apply the Chain Rule

Apply the chain rule which states that \( f'(x) = g'(u(x)) \cdot u'(x) \). Substituting the derivatives, we have:\[ f'(x) = \frac{1}{3}(2x + (x^2 + x)^4)^{-2/3} \cdot \left(2 + 4(x^2 + x)^3(2x + 1)\right) \]

Step 5: Simplify the Expression

Although simplification depends on the context needed, let's tidy the expression:\[ f'(x) = \frac{1}{3}\left(2x + (x^2 + x)^4\right)^{-2/3}(2 + 4(x^2 + x)^3(2x + 1)) \]. Now, the derivative is a bit more concise, capturing the necessary differentiation within one expression.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. At its core, differentiation requires us to compute the derivative of a function, which provides critical insight into how the function behaves.
The process involves several rules and techniques, with the Chain Rule being particularly useful for tackling more complex functions. Here's a simple breakdown of what differentiation encompasses:

• The derivative of a function represents its slope at any particular point, essentially showing how steeply it rises or falls.
• Through differentiation, we can determine important features of functions, such as turning points, maximum and minimum values, and regions where the function is increasing or decreasing.
• Differentiation uses rules like the power rule, product rule, quotient rule, and chain rule to simplify finding derivatives of various types of functions.

In practice, to differentiate functions like the one given in the exercise, the process often requires identifying simpler inner and outer functions, determining their derivatives, and then applying combinations of rules, as demonstrated by the chain rule.

Composite Function

A composite function involves two or more functions combined in such a way that the output of one function becomes the input of another. This composition can make functions appear complicated, but it is essential in many applications of calculus.
In the exercise, the function is formed by nesting two separate functions:

• The inner function, denoted as \( u(x) = 2x + (x^2 + x)^4 \), computes a result which is then subjected to another operation by the outer function.
• The outer function is \( g(u) = u^{1/3} \), which takes the output of the inner function and applies the cube root.

Understanding composite functions helps clarify the structure of complex expressions and is critical when applying the chain rule.
By decomposing the function into its components, you can handle each piece individually when differentiating, ultimately using their derivatives to find the derivative of the overall composite function.

Calculus

Calculus is a powerful field of mathematics that deals with change and motion. It includes differentiation (as discussed) and integration. These processes have far-reaching applications in various disciplines such as physics, engineering, economics, and biology.
In calculus, differentiation is viewed as a method to compute instantaneous rates of change, while integration deals with the accumulation of quantities, essentially finding areas under curves.

• The key tools of calculus include limits, derivatives, and integrals.
• Limits help in understanding the behavior of functions as inputs approach certain values, forming the foundation of both derivatives and integrals.
• Calculus uses strategies and rules like the chain rule to tackle functions that are otherwise difficult to handle with basic algebraic methods.

The exercise demonstrates the application of differential calculus through the chain rule, showcasing how detailed processes can unravel the complexities of mathematically intricate problems.