Problem 68 - Calculus and its applications

Question

Find the first through the fourth derivatives. Be sure to simplify each derivative before continuing. $$ f(x)=\frac{x+3}{x-2} $$

Step-by-Step Solution

Answer

The first derivative is \(\frac{-5}{(x-2)^2}\), second is \(\frac{10}{(x-2)^3}\), third is \(\frac{-30}{(x-2)^4}\), and fourth is \(\frac{120}{(x-2)^5}\)."

Step 1: Find the First Derivative

To find the first derivative of the function \( f(x) = \frac{x+3}{x-2} \), we'll use the quotient rule. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = x+3 \) and \( v(x) = x-2 \). First, let's find \( u'(x) \) and \( v'(x) \). \( u'(x) = 1 \) and \( v'(x) = 1 \). Substituting into the quotient rule, we get:\[ f'(x) = \frac{(1)(x-2) - (x+3)(1)}{(x-2)^2} = \frac{x-2-x-3}{(x-2)^2} \]Simplify the numerator:\[ f'(x) = \frac{-5}{(x-2)^2} \]

Step 2: Find the Second Derivative

Now, we need to find the derivative of \( f'(x) = \frac{-5}{(x-2)^2} \). To simplify, rewrite it as \( f'(x) = -5(x-2)^{-2} \), and use the power rule. The power rule states: if \( h(x) = ax^n \), then \( h'(x) = anx^{n-1} \). Apply this:\[ f''(x) = -5 \times (-2)(x-2)^{-3} \]Simplifying further gives:\[ f''(x) = 10(x-2)^{-3} = \frac{10}{(x-2)^3} \]

Step 3: Find the Third Derivative

Next, find the derivative of \( f''(x) = \frac{10}{(x-2)^3} \). Rewrite it as \( f''(x) = 10(x-2)^{-3} \) and differentiate using the power rule:\[ f'''(x) = 10 \times (-3)(x-2)^{-4} \]Simplify:\[ f'''(x) = -30(x-2)^{-4} = \frac{-30}{(x-2)^4} \]

Step 4: Find the Fourth Derivative

Finally, find the derivative of \( f'''(x) = \frac{-30}{(x-2)^4} \). Rewrite it as \( f'''(x) = -30(x-2)^{-4} \), and use the power rule:\[ f''''(x) = -30 \times (-4)(x-2)^{-5} \]Simplify:\[ f''''(x) = 120(x-2)^{-5} = \frac{120}{(x-2)^5} \]

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus used to find the derivative of a function that is the ratio of two differentiable functions. If we have a function expressed as a fraction like \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule helps us calculate its derivative efficiently.

The rule states:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

This might look a bit complex, but let's break it down:

• Start by finding the derivatives of the top \( u'(x) \) and bottom \( v'(x) \) parts of the fraction.
• Then, multiply the derivative of the top \( u'(x) \) by the original bottom \( v(x) \), and subtract the product of the original top \( u(x) \) with the derivative of the bottom \( v'(x) \).
• Finally, divide the entire expression by the square of the denominator \([v(x)]^2\).

Make sure to simplify the expression at the end for a clearer and simpler derivative. This approach was applied in the original exercise to differentiate the function \( f(x) = \frac{x+3}{x-2} \), resulting in the first derivative \( f'(x) = \frac{-5}{(x-2)^2} \).

Quotient rule may seem tricky initially, but with practice, it becomes an invaluable tool for handling functions involving division.

Power Rule

The power rule is one of the simplest and most frequently used techniques in calculus for differentiating functions. It's particularly useful when dealing with polynomials or functions expressed as powers of \( x \). The rule is straightforward:

• If \( h(x) = ax^n \), then \( h'(x) = anx^{n-1} \).

In simple terms, it says: "Multiply the exponent \( n \) by the coefficient \( a \), and then reduce the exponent by one."

This rule is not just applicable to basic polynomials but also to more complex expressions that can be rewritten in terms of powers, like fractions or roots.
The original exercise applied this rule by rewriting the derivative expressions in terms of negative powers. For example, the first derivative \( f'(x) = \frac{-5}{(x-2)^2} \) was rewritten as \( f'(x) = -5(x-2)^{-2} \) to use the power rule effectively.

Each subsequent derivative was then found using the power rule, leading to increasingly higher powers of negative exponents, which were simplified back into fraction form for clarity.

The power rule is effective, intuitive, and a vital part of calculus problem-solving.

Higher Derivatives

Higher derivatives refer to the derivatives of a derivative. In other words, it's the process of taking the derivative of a function multiple times. This concept is essential when analyzing the behavior of functions beyond simple change of rate, such as acceleration in physics.

In the original exercise, after finding the first derivative of \( f(x) = \frac{x+3}{x-2} \), the task was to continue differentiating to reach up to the fourth derivative. Here's a quick breakdown:

• The first derivative \( f'(x) \) gives us the rate of change of \( f(x) \).
• The second derivative \( f''(x) \) can indicate the concavity of \( f(x) \), helping discern whether a function is concaving up or down.
• The third derivative \( f'''(x) \), though less common, finds use in specific applications like jerk in physics.
• The fourth derivative \( f^{(4)}(x) \) extends this further, usually for theoretical exploration rather than practical scenarios.

Calculating higher derivatives often requires repeated application of rules like the power rule and quotient rule, as seen in our example. Make sure to simplify each derivative before proceeding to maintain clarity and accuracy during these calculations.

Higher derivatives can unveil deep insights into the behavior of mathematical and physical systems.