Problem 74 - Calculus Single Variable

Question

Find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=\left(2 x^{2}+1\right)^{1 / 3}\left(3 x^{3}+1\right)^{4 / 3} $$

Step-by-Step Solution

Answer

\( f'(x) = \frac{4x(3x^3+1)^{4/3}}{3(2x^2+1)^{2/3}} + \frac{12x^2(2x^2+1)^{1/3}}{(3x^3+1)^{2/3}} \)

Step 1: Identify the Functions

Given the function \( f(x) = \left(2x^{2} + 1\right)^{1/3}\left(3x^{3} + 1\right)^{4/3} \), identify \( u(x) = \left(2x^{2} + 1\right)^{1/3} \) and \( v(x) = \left(3x^{3} + 1\right)^{4/3} \). We will need the derivatives of these functions for the product rule.

Step 2: Differentiate each function

First, differentiate \( u(x) = (2x^2 + 1)^{1/3} \) using the chain rule:\[ u'(x) = \frac{1}{3}(2x^2 + 1)^{-2/3} \cdot 4x = \frac{4x}{3(2x^2 + 1)^{2/3}} \]Next, differentiate \( v(x) = (3x^3 + 1)^{4/3} \) using the chain rule:\[ v'(x) = \frac{4}{3}(3x^3 + 1)^{1/3} \cdot 9x^2 = \frac{12x^2}{(3x^3 + 1)^{2/3}} \]

Step 3: Apply the Product Rule

Use the product rule to differentiate \( f(x) = u(x)v(x) \). The product rule is given by:\[f'(x) = u'(x)v(x) + u(x)v'(x) \]Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the product rule:\[f'(x) = \left(\frac{4x}{3(2x^2+1)^{2/3}}\right)\left((3x^3+1)^{4/3}\right) + \left((2x^2+1)^{1/3}\right)\left(\frac{12x^2}{(3x^3+1)^{2/3}}\right) \]

Step 4: Simplify the Expression

Simplify by factoring out common terms if possible, and simplify the fractions:\[f'(x) = \frac{4x(3x^3+1)^{4/3}}{3(2x^2+1)^{2/3}} + \frac{12x^2(2x^2+1)^{1/3}}{(3x^3+1)^{2/3}} \]This expression can be left as an exact representation of the derivative of \( f(x) \).

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate the product of two functions. It is especially useful when you're dealing with expressions that involve the multiplication of separate functions of the variable. The product rule is mathematically stated as:\[(fg)' = f'g + fg'\]Here, if you have two functions, \(f(x)\) and \(g(x)\), whose product is being differentiated, the rule implies the process involves taking the derivative of the first function and multiplying it by the second function. Additionally, you'll also take the derivative of the second function and multiply it by the first one.

• Start with two identifiable functions that compose a larger function to be differentiated.
• Apply the rule meticulously, as each function's derivative contributes to the final result.
• Keep a lookout for simplifying opportunities after applying the product rule to tidy the expression.

In the given problem, the solution involved identifying \(u(x)\) and \(v(x)\) and then using the product rule to find \(f'(x)\) successfully.

Chain Rule

The chain rule is a crucial tool for differentiating compositions of functions. It enables us to work through more intricate expressions where one function is nested inside another.In essence, the chain rule in calculus is represented as:\[(f(g(x)))' = f'(g(x)) \, g'(x)\]This means that when you need the derivative of a function within a function, you take the derivative of the outer function first, leaving the inner function unchanged. Then you multiply it by the derivative of the inner function.Here’s how it helps:

• Break down nested functions into manageable components.
• First, tackle the derivative of the outer function, considering the inner function as a single entity.
• Next, differentiate the inner function.

In the original exercise, the chain rule was applied separately to \(u(x) = (2x^2 + 1)^{1/3}\) and \(v(x) = (3x^3 + 1)^{4/3}\), easing the differentiation process of these composite functions. Understanding each part is imperative to solve joint functions correctly.

Simplifying Derivatives

Simplifying derivatives is an important part of differentiating functions, aiming to provide a clearer and often more usable form of the expression.After applying differentiation techniques, such as the product and chain rules, we often end up with complex expressions. Simplification involves reducing the result to a more concise form.Benefits of simplifying derivatives:

• Clarifies the final result, making it easier to interpret.
• Reduces possibility of errors in further calculations.
• Facilitates a better understanding of the mathematical relationships involved.

In our problem, simplifying was integral after applying the product rule:- We simplified by combining terms and eliminating unnecessary factors, leading to a neat and easy-to-read final expression for \(f'(x)\).It's always a valuable skill to revisit your derivative and tidy up wherever possible. This not only makes your answers look cleaner but ensures accuracy.