Problem 22 - Calculus and its applications

Question

Find \(f^{\prime \prime}(x)\) $$ f(x)=\left(2 x^{2}-3 x+1\right)^{10} $$

Step-by-Step Solution

Answer

The second derivative is \(f''(x) = 10 (2x^2-3x+1)^8 (152x^2 - 228x + 85).\)

Step 1: First Derivative

Begin by finding the first derivative of the function \(f(x)=\left(2x^2-3x+1\right)^{10}\) using the chain rule. Let \(u=2x^2-3x+1\), so \(f(x)=u^{10}\). Applying the chain rule, we have \(f'(x) = 10u^9 \cdot u'\).Now, compute \(u'\):\[u' = \frac{d}{dx}(2x^2-3x+1) = 4x-3.\]Thus, \[f'(x) = 10 \cdot (2x^2-3x+1)^9 \cdot (4x-3).\]

Step 2: Simplify First Derivative

Simplify the expression for \(f'(x)\):\[f'(x) = 10(4x-3)(2x^2-3x+1)^9.\] Keep this expression handy as it will be used in the following step.

Step 3: Second Derivative Using the Product Rule

To find \(f''(x)\), apply the product rule to \(f'(x) = 10(4x-3)(2x^2-3x+1)^9\), which states that if \(f(x) = g(x)h(x)\), then \(f''(x) = g''(x)h(x) + 2g'(x)h'(x) + g(x)h''(x)\).Here, let\(g(x) = 4x-3\) and \(h(x) = (2x^2-3x+1)^9\).We have \(g'(x) = 4\) and \(h'(x)\) can be found using the chain rule as before, but this time for the derivative of \((2x^2-3x+1)^{9}\):Let \(v=(2x^2-3x+1)\), then \(h(x)=v^9\) and \(h'(x)=9v^8v'\), with \(v'=4x-3.\)So,\[h'(x) = 9(2x^2-3x+1)^8(4x-3).\]Thus applying the product rule:\[f''(x) = 10\left[4 \cdot (2x^2-3x+1)^9 + (4x-3) \cdot 9(2x^2-3x+1)^8(4x-3)\right].\]

Step 4: Simplify Second Derivative

Simplify the expression for \(f''(x)\):\[f''(x) = 10 \left[4 (2x^2-3x+1)^9 + 9(4x-3)^2(2x^2-3x+1)^8\right]\]Factor out \((2x^2-3x+1)^8\):\[f''(x) = 10 (2x^2-3x+1)^8 \cdot \left[ 4 (2x^2-3x+1) + 9 (4x-3)^2 \right].\]To simplify further, expand the polynomial:1. Expand \(9(4x-3)^2 = 9(16x^2 - 24x + 9) = 144x^2 - 216x + 81\)Then:\[4 (2x^2-3x+1) = 8x^2 - 12x + 4\]Combine and simplify:\[8x^2 - 12x + 4 + 144x^2 - 216x + 81 = 152x^2 - 228x + 85\]Finally:\[f''(x) = 10 (2x^2-3x+1)^8 (152x^2 - 228x + 85).\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. Imagine you have a function inside another function, like peeling an onion layer by layer. In essence, if you have a function of the form \( f(g(x)) \), the chain rule helps you differentiate it by focusing on both functions separately.
To apply the chain rule, you first differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function. For our original problem, the function \( f(x) = (2x^2 - 3x + 1)^{10} \) can be seen as an outer function of \( u^{10} \), where \( u = 2x^2 - 3x + 1 \).
For easy differentiation:

• Differentiate the outer function: \( \frac{d}{du}u^{10} = 10u^9 \).
• Compute the derivative of the inner function: \( \frac{d}{dx}(2x^2-3x+1) = 4x-3 \).
• Now multiply the results: \( 10u^9 \cdot (4x-3) \).

This combination of differentiating both layers is the essence of the chain rule, simplifying the process when tackling composite functions.

Product Rule

When a function is expressed as the product of two functions, the product rule becomes indispensable to differentiate it correctly. It's like baking a complex cake from two separate recipes!
In our problem when finding the second derivative \( f''(x) \), we used the product rule to differentiate \( f'(x) = 10(4x-3)(2x^2-3x+1)^9 \).

• For the product rule, remember: if \( f(x) = g(x) \cdot h(x) \), then \( f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \).
• Here, assign \( g(x) = 4x-3 \) and \( h(x) = (2x^2-3x+1)^9 \).
• Differentiate each: \( g'(x) = 4 \) and apply the chain rule to find \( h'(x) = 9(2x^2 - 3x + 1)^8 \cdot (4x - 3) \).
• Substitute into the product rule formula to find \( f''(x) \).

The product rule is particularly helpful when dealing with complex products, giving a straightforward method to follow.

Polynomial Simplification

Polynomial simplification involves reducing a polynomial expression into its simplest form. Imagine cleaning up a messy room to make it look neat!
In our task, we simplified the expression for the second derivative \( f''(x) \) to make it more elegant and manageable. The main steps to polynomial simplification include:

• Identifying like terms: Combine terms that have the same degree.
• Expanding expressions: For instance, \( 9(4x-3)^2 = 9(16x^2 - 24x + 9) \) expands to \( 144x^2 - 216x + 81 \).
• Summing coefficients: Add the coefficients of like terms, such as \( 8x^2 \) and \( 144x^2 \) to get \( 152x^2 \).

Once you've expanded and combined all the parts, you achieve a simplified version that is easier to work with, especially when checking or using in further calculations.