Problem 45 - Calculus - AP Edition

Question

Find (a) \(f^{\prime \prime \prime}(2),\) where \(f(x)=3 x^{2}-2\) (b) \(\left.\frac{d^{2} y}{d x^{2}}\right|_{x=1},\) where \(y=6 x^{5}-4 x^{2}\) (c) \(\left.\frac{d^{4}}{d x^{4}}\left[x^{-3}\right]\right|_{x=1}\)

Step-by-Step Solution

Answer

(a) 0, (b) 112, (c) 360.

Step 1: Calculate first, second, and third derivative for f(x)

Given \( f(x) = 3x^2 - 2 \).1. First derivative, \( f'(x) \), is obtained by differentiating \( f(x) \): \[ f'(x) = \frac{d}{dx}(3x^2 - 2) = 6x. \]2. Second derivative, \( f''(x) \): \[ f''(x) = \frac{d}{dx}(6x) = 6. \]3. Third derivative, \( f'''(x) \): \[ f'''(x) = \frac{d}{dx}(6) = 0. \] Thus, \( f'''(2) = 0 \).

Step 2: Calculate second derivative of y at x=1

Given \( y = 6x^5 - 4x^2 \).1. First derivative, \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{d}{dx}(6x^5 - 4x^2) = 30x^4 - 8x. \]2. Second derivative, \( \frac{d^2y}{dx^2} \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(30x^4 - 8x) = 120x^3 - 8. \]3. Evaluate at \( x = 1 \): \[ \left. \frac{d^2y}{dx^2} \right|_{x=1} = 120(1)^3 - 8 = 112. \]

Step 3: Find the fourth derivative of \( x^{-3} \) at x=1

Given function is \( x^{-3} \).1. First derivative: \[ \frac{d}{dx}(x^{-3}) = -3x^{-4}. \]2. Second derivative: \[ \frac{d^2}{dx^2}(x^{-3}) = 12x^{-5}. \]3. Third derivative: \[ \frac{d^3}{dx^3}(x^{-3}) = -60x^{-6}. \]4. Fourth derivative: \[ \frac{d^4}{dx^4}(x^{-3}) = 360x^{-7}. \]5. Evaluate at \( x = 1 \): \[ \left. \frac{d^4}{dx^4}(x^{-3}) \right|_{x=1} = 360(1)^{-7} = 360. \]

Key Concepts

Second Derivative

The second derivative provides insight into the curvature or concavity of a function, which indicates whether it is curving up or down. When calculating the second derivative, you're essentially finding how the rate of change of a function's slope itself is changing. This could reveal points where the function changes from concave up to concave down or vice versa.

In the given exercise, for part (b), we calculated the second derivative of the function \( y = 6x^5 - 4x^2 \) at \( x = 1 \). Here are the steps to find it:

• Calculate the first derivative, \( \frac{dy}{dx} = 30x^4 - 8x \).
• Then, find the second derivative by differentiating again, \( \frac{d^2y}{dx^2} = 120x^3 - 8 \).
• Finally, evaluate this result at \( x = 1 \), leading to \( 120 \times 1^3 - 8 = 112 \).

This calculation indicates that when \( x = 1 \), the function is experiencing some degree of upward curvature, with the second derivative value of 112 at this point suggesting a convex curve.

Third Derivative

The third derivative of a function can be thought of as measuring the rate at which the second derivative is changing. In more intuitive terms, it is often related to the concept of "jerk" in physics, describing how the curvature or concavity is itself changing.

In the exercise, for part (a), the function \( f(x) = 3x^2 - 2 \) has a third derivative that is very straightforward:

• The first derivative \( f'(x) = 6x \).
• The second derivative \( f''(x) = 6 \), which is a constant.
• The third derivative is simply \( f'''(x) = 0 \).

This tells us something quite particular: Since the third derivative is zero, the change in curvature is constant at each point along the function \( f(x) = 3x^2 - 2 \). Therefore, the function's rate of change of its slope doesn’t change, indicating consistency in its shape.

Fourth Derivative

The fourth derivative has applications in mathematical contexts dealing with approximations and physical systems, particularly in areas like physics and engineering. When you continue differentiating, each new derivative's meaning becomes more abstract and less intuitive in everyday scenarios, yet they're still crucial in various analyses.

In the exercise, in part (c), we find the fourth derivative of the function \( x^{-3} \) at \( x = 1 \):

• First derivative: \( -3x^{-4} \).
• Second derivative: \( 12x^{-5} \).
• Third derivative: \( -60x^{-6} \).
• Fourth derivative: \( 360x^{-7} \).

When evaluated at \( x = 1 \), the fourth derivative yields \( 360 \). This final result tells us about the rate at which the third derivative changes. In practical terms, it indicates how the evolution of the function’s shape continues at higher orders of differentiation.