Problem 15 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ f(x)=\sqrt{x^{2}-5} \sqrt[3]{x^{2}+3} $$

Step-by-Step Solution

Answer

The derivative is \[ f'(x) = \left( \frac{x}{(x^2 - 5)^{1/2}} \right) (x^2 + 3)^{1/3} + (x^2 - 5)^{1/2} \left( \frac{2x}{3 (x^2 + 3)^{2/3}} \right) \]

Step 1: Rewrite the function

Express the function in terms of exponents for easier differentiation. \( f(x) = (x^2 - 5)^{1/2} (x^2 + 3)^{1/3} \)

Step 2: Apply the product rule

Using the product rule \( (uv)' = u'v + uv' \), where \( u = (x^2 - 5)^{1/2} \) and \( v = (x^2 + 3)^{1/3} \). Calculate \( u' \) and \( v' \).

Step 3: Differentiate \( u \)

Differentiate \( u = (x^2 - 5)^{1/2} \) using the chain rule: \( u' = \frac{1}{2} (x^2 - 5)^{-1/2} \cdot 2x = \frac{x}{(x^2 - 5)^{1/2}} \)

Step 4: Differentiate \( v \)

Differentiate \( v = (x^2 + 3)^{1/3} \) using the chain rule: \( v' = \frac{1}{3} (x^2 + 3)^{-1/3} \cdot 2x = \frac{2x}{3 (x^2 + 3)^{2/3}} \)

Step 5: Apply the product rule

Combine the results using the product rule: \( f'(x) = u'v + uv' \) Substitute \( u, u', v, \) and \( v' \): \[ f'(x) = \left( \frac{x}{(x^2 - 5)^{1/2}} \right) (x^2 + 3)^{1/3} + (x^2 - 5)^{1/2} \left( \frac{2x}{3 (x^2 + 3)^{2/3}} \right) \]

Key Concepts

Product Rule

The product rule is a crucial concept in calculus used to find the derivative of the product of two functions. Simply put, if you have a function that is a product of two functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of the product is given by:
\[ (uv)' = u'v + uv' \]
In our exercise, we applied the product rule to the function \( f(x) = (x^2 - 5)^{1/2} (x^2 + 3)^{1/3} \). Here, we treated \( (x^2 - 5)^{1/2} \) as \( u \) and \( (x^2 + 3)^{1/3} \) as \( v \).
The product rule allowed us to break down the differentiation task into smaller, more manageable parts by finding the derivatives of \( u \) and \( v \) separately and then combining them. This simplification is what makes the product rule such a powerful tool in calculus.

Chain Rule

The chain rule is indispensable when dealing with composite functions, where one function is nested inside another. It helps in finding the derivative of such functions by breaking them down into simpler parts.
In general, if you have a composite function \( f(g(x)) \), the chain rule states that its derivative is:
\[ (f(g(x)))' = f'(g(x)) \times g'(x) \]
In our given exercise, we applied the chain rule to differentiate \( u = (x^2 - 5)^{1/2} \) and \( v = (x^2 + 3)^{1/3} \).
For \( u \), we set it up as an outer function and an inner function:
\[ u = (x^2 - 5)^{1/2} \] Here, \( f(y) = y^{1/2} \) and \( y = x^2 - 5 \). Thus, using the chain rule, we get: \[ u' = \frac{1}{2} (x^2 - 5)^{-1/2} \times 2x = \frac{x}{(x^2 - 5)^{1/2}} \]
We did something similar for \( v = (x^2 + 3)^{1/3} \) by taking the outer function as \( f(y) = y^{1/3} \) and the inner function as \( y = x^2 + 3 \), giving us: \[ v' = \frac{1}{3} (x^2 + 3)^{-1/3} \times 2x = \frac{2x}{3 (x^2 + 3)^{2/3}} \]
The chain rule's versatility in handling composite functions makes it one of the fundamental tools in differentiation.

Differentiation

Differentiation is one of the main operations in calculus, focused on finding the rate at which a function changes at any given point. It involves calculating the derivative, which measures the slope of the function's graph.
In our exercise, we wanted to find the derivative of:
\[ f(x) = \sqrt{x^2 - 5} \sqrt[3]{x^2 + 3} \]
The overall process of differentiation in this context involved several steps:
\[ 1. Rewriting the function using exponents for simplicity:
f(x) = (x^2 - 5)^{1/2} (x^2 + 3)^{1/3} \] 2. Applying the product rule to break down the differentiation process.
3. Using the chain rule to differentiate inner and outer functions.
4. Combining the results of the individual derivatives to get:
\[ f'(x) = \frac{x}{(x^2 - 5)^{1/2}} (x^2 + 3)^{1/3} + (x^2 - 5)^{1/2} \frac{2x}{3 (x^2 + 3)^{2/3}} \]
Every step of this process relied on understanding and applying fundamental differentiation rules, emphasizing how these rules interplay to solve complex problems effectively.