Problem 20 - Calculus an Applied Approach

Question

find the third derivative of the function. $$ f(x)=\left(x^{3}-6\right)^{4} $$

Step-by-Step Solution

Answer

The third derivative of the function \( f(x) = (x^{3}-6)^{4} \) is \( f'''(x) = 36(x^{3}-6)^{2} \cdot (30x^{4} + 28x^{2} - 3) \).

Step 1: Find the First Derivative

Differentiate \( f(x) = (x^{3}-6)^{4} \) using the chain rule. The chain rule, in this case, would be applied as follows: the derivative of the outside function, keeping the inside function the same, multiplied by the derivative of the inside function. This gives \( f'(x) = 4(x^{3}-6)^{3} \cdot 3x^{2} \).

Step 2: Find the Second Derivative

For \( f''(x) \), differentiate the first derivative \( f'(x) = 12x^{2}(x^{3}-6)^{3} \) using the chain rule and product rule. Here, let \( u=x^{3}-6 \) and \( v=x^{2} \). Then \( f''(x) \) becomes \( f''(x) = 36x(x^{3}-6)^{2}\cdot(x^{2}+4x^{4}) \).

Step 3: Find the Third Derivative

Finally, for the third derivative, \( f'''(x) \), differentiate the second derivative \( f''(x) = 36x(x^{3}-6)^{2}\cdot(x^{2}+4x^{4}) \) using the chain rule and product rule once again. This gives \( f'''(x) = 36(x^{3}-6)^{2} \cdot (30x^{4} + 28x^{2} - 3) \).

Key Concepts

Chain Rule

The chain rule is an essential tool for differentiating composite functions. Composite functions are functions nested within each other, like in the function given: \( f(x) = (x^3 - 6)^4 \). If we consider this as a function \(g\) inside another function \(h\), then \(g(x) = x^3 - 6\) and \(h(u) = u^4\), where \(u = g(x)\).
The chain rule states that the derivative of \(h(g(x))\) is the derivative of \(h(u)\) with respect to \(u\), times the derivative of \(g(x)\) with respect to \(x\). This looks like:

• Derivative of the outer function: \(h'(u) = 4u^3\)
• Derivative of the inner function: \(g'(x) = 3x^2\)
• Apply the chain rule: \(f'(x) = 4(x^3 - 6)^3 \, (3x^2)\)

The chain rule helps simplify the process of handling and differentiating these complicated functions with nested layers.

Product Rule

The product rule is employed when differentiating functions that are products of two separate functions. For example, when finding the second derivative in the given problem, we encountered \(f'(x) = 12x^2(x^3 - 6)^3\).
In this scenario, think of these as two functions, \(u\) and \(v\), where \(u(x) = 12x^2\) and \(v(x) = (x^3 - 6)^3\). The product rule states that to differentiate \(u(x)v(x)\), you do:

• \( (uv)' = u'v + uv' \)
• \( u'(x) = 24x \)
• \( v'(x) = 3(x^3 - 6)^2 \, (3x^2) \)

Using the product rule, this simplifies to: \[ f''(x) = 24x(x^3 - 6)^3 + 12x^2 \cdot 3(x^3 - 6)^2 \cdot 3x^2 \]The product rule ensures each part of the product is considered and accurately differentiated, maintaining the integrity of the function.

Higher-Order Derivatives

Higher-order derivatives involve finding the second, third, or even higher derivatives of a function. These derivatives reveal information about the behavior and curvature of the function at various points.
In the given exercise, the third derivative \(f'''(x)\) is found by differentiating the second derivative \(f''(x) = 36x(x^3 - 6)^2 \cdot (x^2 + 4x^4)\). In this step:

• Carefully apply both the product and chain rules again.
• Ensure each differentiation process is followed correctly.
• The result gives the third derivative: \( f'''(x) = 36(x^3-6)^2 \cdot (30x^4 + 28x^2 - 3) \)

Having the third derivative helps understand how the rate of change of the rate of change (acceleration, for instance) behaves. It’s particularly valuable in physics and dynamics studies where such details are crucial.