Problem 28 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=\frac{1+z}{\ln z} $$

Step-by-Step Solution

Answer

The derivative is \(\frac{z\ln z - 1 - z}{z(\ln z)^2}\).

Step 1: Identification of Differentiation Technique

The given function is a rational function where both the numerator and the denominator are functions of the variable \(z\). Therefore, we need to apply the quotient rule for differentiation, which is \(\frac{d}{dz}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\). Here, \(u = 1 + z\) and \(v = \ln z\).

Step 2: Differentiate the Numerator

Differentiate the numerator \(u = 1 + z\) with respect to \(z\). The derivative of \(1 + z\) is simply \(1\), so \(u' = 1\).

Step 3: Differentiate the Denominator

Differentiate the denominator \(v = \ln z\) with respect to \(z\). The derivative of \(\ln z\) is \(\frac{1}{z}\), so \(v' = \frac{1}{z}\).

Step 4: Apply the Quotient Rule

Substitute the derivatives obtained in Steps 2 and 3 into the quotient rule formula. We have:\[\frac{d}{dz}\left(\frac{1+z}{\ln z}\right) = \frac{(1)(\ln z) - (1+z)\left(\frac{1}{z}\right)}{(\ln z)^2}.\]

Step 5: Simplify the Expression

Simplify the expression obtained in Step 4:\[= \frac{\ln z - \frac{1+z}{z}}{(\ln z)^2} = \frac{z\ln z - (1+z)}{z(\ln z)^2}.\]This simplifies to:\[ = \frac{z\ln z - 1 - z}{z(\ln z)^2}.\]

Key Concepts

Quotient Rule

When dealing with functions expressed as a fraction, where one function is divided by another, the quotient rule is an essential differentiation tool. It helps us find the derivative of such rational functions. The rule is expressed as:

• \[ \frac{d}{dz}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

Here, \( u \) represents the numerator of the function, while \( v \) corresponds to the denominator. Applying this rule requires us to determine the derivatives of both \( u \) and \( v \) separately before we substitute back into the formula. It ensures that we properly account for the interconnectedness of the numerator and denominator's variations with respect to changes in the variable \( z \). This technique is particularly useful because it systematically deals with the complexities introduced by the division of two functions.

Rational Function

A rational function is a function that can be expressed as the ratio of two polynomial functions. In mathematical notation, it takes the form:

• \[ y = \frac{f(x)}{g(x)} \]

where both \( f(x) \) and \( g(x) \) are polynomials, and \( g(x) eq 0 \). In the exercise, the given function is:

• \[ y = \frac{1+z}{\ln z} \]

Although \( \ln z \) is not a polynomial, the function still behaves similarly because it is expressed as the ratio of two functions with a clear numerator and denominator. Understanding rational functions is crucial because they often show up in calculus problems. Deriving them requires caution as we need to apply specific rules like the quotient rule to manage the division aspect effectively.

Differentiation Technique

Differentiation techniques allow us to find the rate of change of a function concerning its variable. In calculus, there are several techniques available depending on the function's setup. For our exercise, the function requires the application of the quotient rule because it is in the form \( \frac{u}{v} \).
To summarize the process:

• Identify the type of function. Is it a product, quotient, or chain of functions?
• Choose the appropriate rule or technique. Here, we use the quotient rule because of the division between two functions.
• Calculate the derivatives of the individual components, \( u \) and \( v \) in this case.
• Apply these derivatives in the rule to find the overall derivative. This often requires careful algebraic manipulation to simplify.

These techniques simplify otherwise complex calculations, and understanding them provides a solid foundation for tackling diverse calculus problems.