Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate. Spend time writing each of these in a form that makes the differentiation easy. (a) \(y=3 \ln 5 x+6 \ln \left(\frac{3}{x}\right)\) (b) \(y=20 \log \left(\frac{x}{100}\right)\) (c) \(y=\frac{k e^{2 k x}}{\sqrt{k+1}}\) (d) \(y=\left(\frac{(\ln 2) e^{5 x}}{\ln 4}\right)+\frac{(\ln 2) e^{\ln 2}}{\ln 3}\)

Step-by-Step Solution

Answer

The derivatives are (a) \( \frac{3}{x} \), (b) \( \frac{20}{x} \), (c) \( 2k^2 e^{2 k x} (k + 1)^{-\frac{1}{2}} - \frac{1}{2} k e^{2 k x}(k + 1)^{-\frac{3}{2}} \), and (d) \( \frac{5 e^{5x + \ln\ln2}}{\ln4} \)

Step 1: Differentiating function (a)

First simplify the given function applying the logarithm property \( a \ln b = \ln b^a \). So, \( y = ln(5^3 * x^3) \). The derivative is then \( \frac{3}{x} \)

Step 2: Differentiating function (b)

Rewrite the function by using logarithm rules \( \log_b a = \frac{\ln a}{\ln b} \), such that \( y = \frac{20 \ln x}{\ln 100} \). Using chain rule to differentiate, \( y' = \frac{20}{x} \)

Step 3: Differentiating function (c)

First simplify the function by expressing it in exponential form: \( y = k e^{2 k x} (k + 1)^{-\frac{1}{2}} \). Now, apply the product rule and chain rule for differentiation: \( y' = 2k^2 e^{2 k x} (k + 1)^{-\frac{1}{2}} - k e^{2 k x}(k + 1)^{-\frac{3}{2}} \)

Step 4: Differentiating function (d)

Simplify the function applying exponential property \( a^{log_b c} = c^{log_b a} \), such that \( y = \frac{e^{5 x + \ln \ln 2}}{\ln 4} + \frac{(\ln 2) e}{\ln 3} \). By differentiating the simplified function, we get \( y' = \frac{5 e^{5x + \ln\ln2}}{\ln4} \)

Key Concepts

Logarithm Properties

To simplify and differentiate logarithmic functions, it's crucial to be familiar with logarithm properties. For instance, one common property is the power rule, which states that the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the base, or in mathematical terms, \( a \ln b = \ln b^a \). This property allows us to transform complex expressions into simpler forms that make differentiation more straightforward. When you come across expressions like \(y=3 \ln 5x+6 \ln \left(\frac{3}{x}\right)\), using the power rule to consolidate the terms into a single logarithm \(\ln(5^3 * x^3)\) paves the way for a much easier differentiation process.

Another property often used in calculus is the change of base formula, which lets you rewrite logarithms with one base in terms of logarithms with another base, formally written as \( \log_b a = \frac{\ln a}{\ln b} \). Using this property, we can convert non-natural logarithmic expressions to natural logarithms, as seen in exercise (b). These logarithmic properties are the building blocks that lead to easier and more accurate differentiation.

Chain Rule

When differentiating composite functions, the chain rule becomes an indispensable tool in calculus. The chain rule states that to differentiate a composite function \( f(g(x)) \), you must take the derivative of the outer function evaluated at the inner function and then multiply it by the derivative of the inner function. Mathematically, this is denoted as \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).

For example, when differentiating a function like \(y=20 \log \left(\frac{x}{100}\right)\) after applying the change of base formula, the outer function becomes \( 20 \ln x \), and the inner function is the variable \( x \). Using the chain rule facilitates the differentiation of such expressions, yielding a simple result like \( y' = \frac{20}{x} \). It's essential to recognize when the chain rule applies and correctly identify the inner and outer functions for its proper application.

Product Rule

The product rule is a derivative rule used when we have to differentiate an expression that is the multiplication of two or more functions. The formula for the product rule is \( (fg)' = f'g + fg' \), which means the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

In exercise (c), we have a function \(y=\frac{k e^{2 k x}}{\sqrt{k+1}}\), which can be rewritten as \(y=k e^{2 k x}(k + 1)^{-\frac{1}{2}}\). Here, the two functions being multiplied are \(k e^{2 k x}\) and \( (k + 1)^{-\frac{1}{2}}\). Applying the product rule and the chain rule for each part, we derive the expression \( y' = 2k^2 e^{2 k x} (k + 1)^{-\frac{1}{2}} - k e^{2 k x}(k + 1)^{-\frac{3}{2}} \). This demonstrates the importance of applying the correct rules and simplifying terms to facilitate differentiation of products.

Exponential Differentiation

Differentiating exponential functions involves specific rules, especially when the base of the exponent is a constant like \(e\), the natural logarithm base. The differentiation of a natural exponential function \(e^{f(x)}\) is relatively straightforward; the derivative is \(e^{f(x)} \cdot f'(x)\), where \(f'(x)\) is the derivative of the exponent.

For instance, in exercise (d), we initially simplify our function using exponential properties before differentiating. Once we have a function in the form of \(e^{g(x)}\), then applying the derivative rules for exponential functions results in a simple product of the original exponential expression and the derivative of the exponent, like \(y' = \frac{5 e^{5x + \ln\ln2}}{\ln4}\) from the simplified expression of function (d).

Calculus

Calculus is the study of how things change. It provides a framework for modeling systems in which there is change and a way to deduce the predictions of such models. The differentiation process, which includes the various rules and properties discussed, such as the chain rule, product rule, and properties of logarithms and exponentials, is one of the two primary operations in calculus, the other being integration.

Through the differentiation techniques demonstrated in each exercise, calculus allows us to analyze the rates at which quantities change. For instance, the exercises presented involve simplifying complex expressions using algebraic properties before applying calculus concepts, resulting in a step-by-step solution that captures the essence of the calculus approach: breaking down a problem into solvable parts and then synthesizing the information into the desired derivative or predicted outcome.