Problem 26 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the second derivative: a. \(f(x)=6 x^{5}-4 x^{3}+5 x^{2}-2 x+\frac{1}{x}\) b. \(z=\frac{2}{1+x^{2}}\) c. \(y=\left(3 x^{2}+2\right)^{4}\)

Step-by-Step Solution

Answer

a. \[f''(x) = 120x^3 - 24x + 10 + \frac{2}{x^3}\], b. \[z''(x) = \frac{4 + 12x^2}{(1 + x^2)^3}\], c. \[y''(x) = 24(3x^2+2)^3 + 432x^2(3x^2+2)^2\].

Step 1 Title - Differentiate the function for the first time (a)

Find the first derivative of the function. Given function: \(f(x)=6 x^{5}-4 x^{3}+5 x^{2}-2 x+\frac{1}{x}\)The first derivative is found using the power rule \[f'(x)=\frac{d}{dx}\left(6x^5 - 4x^3 + 5x^2 - 2x + \frac{1}{x}\right) = 30x^4 - 12x^2 + 10x - 2 - \frac{1}{x^2}\].

Step 2 Title - Differentiate for the second time (a)

Find the second derivative of the first derivative function:\[f''(x) = \frac{d}{dx}\left(30x^4 - 12x^2 + 10x - 2 - \frac{1}{x^2}\right) = 120x^3 - 24x + 10 + \frac{2}{x^3}\].

Step 3 Title - Differentiate the function for the first time (b)

Given function: \(z=\frac{2}{1+x^{2}}\)Use the quotient rule: \[z'(x) = \frac{d}{dx}\left(\frac{2}{1+x^2}\right) = \frac{(0)(1+x^2) - 2(2x)}{(1+x^2)^2} = -\frac{4x}{(1+x^2)^2}\].

Step 4 Title - Differentiate for the second time (b)

Differentiate the first derivative again:Use the quotient rule and chain rule:\[z''(x) = \frac{d}{dx}\left(-\frac{4x}{(1+x^2)^2}\right) = \frac{(4(1+x^2)^2) -(-4x) (2(1+x^2)2x)}{(1+x^2)^4} = \frac{4(1+x^2)^2 + 32x^2}{(1+x^2)^3}\].

Step 5 Title - Simplify second derivative (b)

Simplify the equation from the previous step:\[z''(x) = \frac{4(1 + x^2) + 8x^2}{(1 + x^2)^3} = \frac{4 + 12x^2}{(1 + x^2)^3}\].

Step 6 Title - Differentiate the function for the first time (c)

Given function: \(y=(3 x^{2}+2)^{4}\)Use the chain rule: \[y'(x) = 4(3 x^{2}+2)^{3} \times \frac{d}{dx}(3x^2+2) = 4(3 x^{2}+2)^{3} \times 6x = 24x(3x^2+2)^{3}\].

Step 7 Title - Differentiate for the second time (c)

Differentiate the first derivative again using the product rule and chain rule:\[y''(x) = \frac{d}{dx}\left(24x(3x^2+2)^{3}\right) = 24(3x^2+2)^3 + 24x \times 3(3x^2+2)^2 \times 6x\].Thus, \[y''(x) = 24(3x^2+2)^3 + 432x^2(3x^2+2)^2\].

Key Concepts

calculus

Calculus is a branch of mathematics focusing on rates of change and the accumulation of quantities. It has two main branches: differential calculus and integral calculus. Differential calculus emphasizes the concept of derivatives, which measure how a function changes as its input changes.
In this exercise, we deal with the second derivative. The second derivative measures the rate at which the first derivative changes. This helps in understanding the concavity of functions and identifying points of inflection.

differentiation

Differentiation is a key concept in differential calculus. It involves finding the derivative of a function, which shows the rate of change of the function's value with respect to changes in its input variable.
The derivative of a function is denoted as \( f'(x) \) or \( \frac{df}{dx} \). For instance, in part (a) of the problem:

• First, we found the first derivative, \( f'(x) = 30x^4 - 12x^2 + 10x - 2 - \frac{1}{x^2} \)
• Then, we found the second derivative, \( f''(x) = 120x^3 - 24x + 10 + \frac{2}{x^3} \).

power rule

The power rule is used to differentiate functions of the form \( x^n \), where \( n \) is a real number. The rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
In part (a), differentiating each term using the power rule:

• For \( 6x^5 \): \( 6 \times 5x^{4} = 30x^{4} \)
• For \( -4x^3 \): \( -4 \times 3x^{2} = -12x^{2} \)
• For \( 5x^2 \): \( 5 \times 2x = 10x \)
• For \( -2x \): \( -2x^0 = -2 \)
• For \( \frac{1}{x} \): rewriting as \( x^{-1} \), \( -x^{-2} = -\frac{1}{x^2} \).

Combining these results gives the first derivative \( f'(x) = 30x^4 - 12x^2 + 10x - 2 - \frac{1}{x^2} \).

quotient rule

The quotient rule is used for differentiating functions that are ratios of two other functions. If \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of \( \frac{u(x)}{v(x)} \) is
\[ \frac{d}{dx}\frac{u(x)}{v(x)} = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \]
For example, in part (b) of the problem:

• The first derivative of \( z=\frac{2}{1+x^{2}} \) is found by applying the quotient rule.
  Here, \( u(x) = 2 \) and \[ v(x) = 1 + x^2 \]
• Applying the quotient rule:
  \[\frac{d}{dx}\frac{2}{1+x^2} = -\frac{4x}{(1+x^2)^2} \]

Then, the second derivative \( z''(x) \) is also found using the quotient rule in combination with the chain rule.

chain rule

The chain rule is used when differentiating compositions of functions. If \( y = g(f(x)) \), then the derivative is \[ \frac{dy}{dx} = g'(f(x)) \times f'(x) \].
For part (c) of the problem, where \( y = (3x^2 + 2)^4 \), we use the chain rule:

• First, let \( u = 3x^2 + 2 \)
• The outer function is \( y = u^4 \)
• Using the chain rule for the first derivative:
  \[ y'(x) = 4(3x^2+2)^3 \times 6x = 24x(3x^2+2)^3 \]
• For the second derivative, we need to apply both the chain rule and the product rule on \[ 24x(3x^2+2)^3 \]:
  \[ y''(x) = \frac{d}{dx}(24x) \times (3x^2+2)^3 + 24x \times \frac{d}{dx}(3x^2+2)^3 \]
• Finally, combining the derivatives:
  \[ y''(x) = 24(3x^2+2)^3 + 432x^2(3x^2+2)^2 \]