Problem 46 - Calculus Early Transcendentals

Question

Find $$ \begin{array}{l}{\text { (a) } y^{\prime \prime \prime}(0), \text { where } y=4 x^{4}+2 x^{3}+3} \\ {\text { (b) }\left.\frac{d^{4} y}{d x^{4}}\right|_{x=1}, \text { where } y=\frac{6}{x^{4}}}\end{array} $$

Step-by-Step Solution

Answer

(a) 12, (b) 5040

Step 1: Differentiate Once

For part (a), the given function is \( y = 4x^4 + 2x^3 + 3 \). The first derivative is found using the power rule: \( y' = \frac{d}{dx}(4x^4 + 2x^3 + 3) = 16x^3 + 6x^2 \).

Step 2: Differentiate Twice

The second derivative is obtained by differentiating the first derivative: \( y'' = \frac{d}{dx}(16x^3 + 6x^2) = 48x^2 + 12x \).

Step 3: Differentiate Thrice

The third derivative is found by differentiating the second derivative: \( y''' = \frac{d}{dx}(48x^2 + 12x) = 96x + 12 \).

Step 4: Evaluate Third Derivative at Zero

Now that we have the third derivative \( y''' = 96x + 12 \), evaluate it at \( x = 0 \): \( y'''(0) = 96(0) + 12 = 12 \).

Step 5: Differentiate Once (Part b)

For part (b), the function is \( y = \frac{6}{x^4} \), which can be rewritten as \( y = 6x^{-4} \). The first derivative is \( y' = \frac{d}{dx}(6x^{-4}) = -24x^{-5} \).

Step 6: Differentiate Twice (Part b)

The second derivative is \( y'' = \frac{d}{dx}(-24x^{-5}) = 120x^{-6} \).

Step 7: Differentiate Thrice (Part b)

The third derivative is \( y''' = \frac{d}{dx}(120x^{-6}) = -720x^{-7} \).

Step 8: Differentiate Four Times (Part b)

The fourth derivative is \( y'''' = \frac{d}{dx}(-720x^{-7}) = 5040x^{-8} \).

Step 9: Evaluate Fourth Derivative at One

Now that we have the fourth derivative \( y'''' = 5040x^{-8} \), evaluate it at \( x = 1 \): \( y''''(1) = 5040(1)^{-8} = 5040 \).

Key Concepts

Third Derivative

The third derivative of a function, represented as \( y''' \), signifies the third rate of change of the original function \( y \). It stems from differentiating the second derivative, which indicates the curvature or concavity of \( y \). Calculating the third derivative can unveil important dynamics about how the function's slope itself changes over time.
In part (a) of the exercise, we started with the function \( y = 4x^4 + 2x^3 + 3 \). Through successive differentiations:

• The first derivative \( y' \) is found to be \( 16x^3 + 6x^2 \), indicating the slope of the tangent line to \( y \).
• The second derivative \( y'' \) is calculated as \( 48x^2 + 12x \), which provides insight into the function's curvature.
• Finally, differentiating one more time, the third derivative \( y''' \) results in \( 96x + 12 \). This derivative reflects the rate of change of the function's curvature.

When evaluated at \( x = 0 \), we find the third derivative yields \( y'''(0) = 12 \). This value tells us about the immediate change in slope at that particular point, showing how the function's rate of change is evolving there.

Fourth Derivative

The fourth derivative, denoted as \( y'''' \), goes one step beyond the third derivative and examines the rate of change of the rate of change's rate of change of \( y \). In simpler terms, it tracks how the "jerk" (the third derivative) itself is changing.
For part (b) of the exercise, the function \( y = \frac{6}{x^4} \) is transformed to \( y = 6x^{-4} \) to make differentiation easier. Through calculating derivatives in succession:

• After finding the first three derivatives, we come to the fourth derivative, \( y'''' = 5040x^{-8} \).

This derivative can sometimes relate to the smoothness of the original function, revealing intricate details about its behavior as we look further into higher orders. When evaluated at \( x = 1 \), the fourth derivative reached the value \( y''''(1) = 5040 \). This number offers a different perspective on the function's behavior, providing insight into the subtler nuances of its movement and changes.

Power Rule

The power rule is a core technique in calculus for differentiating functions of the form \( x^n \). It states that to find the derivative of a function \( f(x) = x^n \), you bring the exponent \( n \) down in front as a coefficient and then reduce the original exponent by 1. This is mathematically expressed as:\[\frac{d}{dx}(x^n) = nx^{n-1}\]In both parts of the exercise, the power rule was pivotal in determining higher-order derivatives.

• For part (a), starting with \( y = 4x^4 + 2x^3 + 3 \), the power rule was applied repeatedly to derive the first, second, and third derivatives, culminating in \( y''' = 96x + 12 \).
• For part (b), the function \( y = \frac{6}{x^4} \) was rewritten as \( y = 6x^{-4} \), allowing for the power rule to be used to efficiently compute derivatives up to the fourth order, concluding with \( y'''' = 5040x^{-8} \).

Understanding and applying the power rule simplifies the process of finding derivatives, making it a fundamental building block in the study of calculus. By consistently applying these rules, we can handle even more intricate functions.