Problem 20 - Calculus and its applications

Question

Find \(f^{\prime \prime}(x)\) $$ f(x)=\left(x^{3}+2 x\right)^{6} $$

Step-by-Step Solution

Answer

The second derivative is \(f''(x) = 6((5(x^3 + 2x)^4(3x^2 + 2)^2) + ((x^3 + 2x)^5 \cdot 6x))\)."

Step 1: Identify the Outer and Inner Functions

We are given the function \(f(x) = (x^3 + 2x)^6\). Here, the outer function is \(u^6\) where \(u = x^3 + 2x\). The inner function is \(x^3 + 2x\).

Step 2: Differentiate the Outer Function

First, differentiate the outer function \(u^6\) with respect to \(u\) using the power rule: \(\frac{d}{du}(u^6) = 6u^5\).

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \(u = x^3 + 2x\) with respect to \(x\): \(\frac{d}{dx}(x^3 + 2x) = 3x^2 + 2\).

Step 4: Apply the Chain Rule for the First Derivative

Using the chain rule, the first derivative \(f'(x)\) is found by multiplying \( \frac{d}{du}(u^6)\) by \( \frac{du}{dx}\): \[f'(x) = 6(x^3 + 2x)^5(3x^2 + 2)\].

Step 5: Differentiate the First Derivative for the Second Derivative

To find the second derivative \(f''(x)\), we must differentiate \(f'(x)\) again with respect to \(x\). This involves using the product rule and chain rule again.

Step 6: Use the Product Rule

The first term is \(6(x^3 + 2x)^5\) and the second term is \(3x^2 + 2\). Differentiating, we apply the product rule: \((u \, v)' = u'v + uv'\).

Step 7: Differentiate the Outer Function in Product Rule

Differentiate the first part of the product: \(6(x^3 + 2x)^5\). Apply the chain rule: \(5 \cdot 6(x^3 + 2x)^4 (3x^2 + 2)\).

Step 8: Differentiate the Second Term in Product Rule

The derivative of \(3x^2 + 2\) is \(6x\).

Step 9: Combine the Derivatives Using Product Rule

Put the derivatives from Steps 7 and 8 into the product rule formula: \[f''(x) = 6((5(x^3 + 2x)^4(3x^2 + 2)^2) + ((x^3 + 2x)^5 \cdot 6x))\].

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus. It allows us to differentiate compositions of functions. When working with complex expressions that are made up of inner and outer functions, the chain rule simplifies the process of differentiation.

To understand how the chain rule works, let's consider the function given in the exercise: \(f(x) = (x^3 + 2x)^6\). We identify the outer function as \(u^6\) and the inner function as \(u = x^3 + 2x\).

When using the chain rule, you derive the outer function and multiply it by the derivative of the inner function:

• Differentiate the outer function \(u^6\) to get \(6u^5\).
• Differentiate the inner function \(x^3 + 2x\) to get \(3x^2 + 2\).

Thus, using the chain rule, the derivative is found by taking \(6u^5\) and substituting \(u\) back as \((x^3 + 2x)\), which results in: \[f'(x) = 6(x^3 + 2x)^5(3x^2 + 2)\].
The chain rule is crucial because it simplifies the task of differentiating composite functions by breaking the function into manageable parts.

Product Rule

The product rule is another essential differentiation technique. Whenever you have two functions being multiplied, the product rule comes into play. This technique helps find the derivative of the product of two functions by considering both functions and their derivatives.

In the exercise, after finding the first derivative using the chain rule, we face the product \(6(x^3 + 2x)^5\) and \(3x^2 + 2\). To differentiate this product, we apply the product rule:

• Let \(u = 6(x^3 + 2x)^5\) and \(v = 3x^2 + 2\).
• The product rule formula is \((uv)' = u'v + uv'\).

First, differentiate each part separately:

• Differentiate \(u = 6(x^3 + 2x)^5\) using the chain rule, giving \(5 \cdot 6(x^3 + 2x)^4(3x^2 + 2)\).
• Differentiate \(v = 3x^2 + 2\) to get \(6x\).

Putting it all together using the product rule gives the second derivative: \[f''(x) = 6\left(5(x^3 + 2x)^4(3x^2 + 2)^2 + (x^3 + 2x)^5 \cdot 6x\right)\].
The product rule allows for orderly and systematic differentiation of function products.

Differentiation Techniques

Differentiation techniques include these rules among others that aid in finding derivatives efficiently. Understanding when and how to apply each rule is key when working through calculus problems.

- **Power Rule**: Used when differentiating terms of the form \(x^n\), where the derivative is \(nx^{n-1}\).
- **Chain Rule**: As discussed, this is critical when functions are nested, helping to break down derivatives of composite functions.
- **Product Rule**: Perfect for derivatives of products of two or more functions. It compartmentalizes the problem into smaller pieces that are individually differentiated and recombined.

Mastering these rules simplifies even the most complex differentiation problems by breaking down calculations into steps that are easier to manage. Practice, alongside theoretical understanding, reinforces these concepts, making solving differentiation problems more intuitive.