Problem 60 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ g(s)=\log _{5}\left(3^{s}-2\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{3^s \ln(3)}{(3^s - 2) \ln(5)} \).

Step 1: Apply the Chain Rule

The function is a composite function, composed of an outer logarithmic function and an inner exponential function. Start by applying the chain rule: if you have a composition of functions, such as \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). Identify the outer function as \( f(u) = \log_5(u) \) where \( u = 3^s - 2 \).

Step 2: Differentiate the Outer Function

Differentiate the outer logarithmic function \( f(u) = \log_5(u) \) with respect to \( u \). The derivative is \( \frac{1}{u \ln(5)} \) since the derivative of \( \log_b(x) \) is \( \frac{1}{x \ln(b)} \).

Step 3: Differentiate the Inner Function

Find the derivative of the inner function \( u = 3^s - 2 \) with respect to \( s \). The derivative of \( 3^s \) with respect to \( s \) is \( 3^s \ln(3) \), since the derivative of \( a^x \) is \( a^x \ln(a) \), and the derivative of a constant (\(-2\)) is zero.

Step 4: Combine Using the Chain Rule

Combine the derivatives from Steps 2 and 3 using the chain rule: \( \frac{dg}{ds} = \frac{1}{u \ln(5)} \cdot 3^s \ln(3) \). Substituting \( u = 3^s - 2 \), we get: \( \frac{dg}{ds} = \frac{3^s \ln(3)}{(3^s - 2) \ln(5)} \).

Key Concepts

Chain Rule

When faced with a complex function that is the combination of two or more functions, like our logarithmic and exponential function, the Chain Rule is a lifesaver. It helps us break down the differentiation of such functions into more manageable parts. The main idea of the Chain Rule is that if you have a composite function like \(f(g(x))\), the derivative is determined by \(f'(g(x)) \, g'(x)\). This rule allows us to first differentiate the outer function, keeping the inner function intact, and then multiply it by the derivative of the inner function. In our specific problem, our composite function is \(\log_5(3^s - 2)\), where the outer function is \(\log_5(u)\) and the inner function is \(3^s - 2\). Using the Chain Rule, we can tackle each piece separately:

• Differentiate the outer function with respect to the inner function.
• Then differentiate the inner function with respect to \(s\).
• Finally, combine these results to get the derivative of the entire composite function.

This staged approach simplifies the differentiation of complex expressions, making the task more approachable.

Logarithmic Differentiation

Logarithmic differentiation is a specialized technique for differentiating functions involving logs. It is particularly useful when dealing with complex expressions or when the base of the logarithm is not the natural base \(e\). While not applied in the usual sense to apply the whole function \(f(x)\), it is still related to our solution, as understanding logarithmic derivatives is crucial. For function \(g(s)=\log_5(3^s-2)\), we specifically use the formula for the derivative of a logarithm: \(\frac{d}{du}\log_b(u) = \frac{1}{u \ln(b)}\). In this exercise:

• Our outer function is \(\log_5(u)\).
• Inevitably, its derivative becomes \(\frac{1}{u \ln(5)}\).

Understanding how to differentiate logs with any base equips us with the skills to tackle a variety of problems efficiently.

Exponential Functions

Exponential functions are functions of the form \(a^x\), where \(a\) is a constant, and \(x\) is the variable. In our exercise, the inner function \(3^s - 2\) features an exponential term \(3^s\). Differentiating such functions involves using the rule that the derivative of \(a^x\) is \(a^x \ln(a)\). Here’s why this is straightforward and handy:

• Recognize \(3^s\) as an exponential function with base 3.
• The derivative follows the rule above, resulting in \(3^s \ln(3)\).

The subtraction of a constant ( -2) within \(3^s - 2\) has a derivative of zero, which streamlines our derivative calculation of the inner function. Comprehending this concept not only helps in current problems but also paves the way for tackling more challenging exponential differentiations in the future.