Problem 60 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ g(s)=\log _{5}\left(3^{s}-2\right) $$

Step-by-Step Solution

Answer

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

Step 1: Recognize the use of the chain rule

To differentiate the function \( g(s) = \log_5 (3^s - 2) \), we will use the chain rule. The chain rule states that if we have a function within another function, say \( h(x) = f(g(x)) \), then the derivative \( h'(x) = f'(g(x)) \cdot g'(x) \). Here, \( f(x) = \log_5(x) \) and \( u(x) = 3^s - 2 \).

Step 2: Differentiate the outer function

The outer function is \( f(u) = \log_5(u) \). The derivative of a logarithm to base 5 is given by \( \frac{1}{u \ln(5)} \) where \( u = 3^s - 2 \). So, \( f'(u) = \frac{1}{(3^s - 2) \ln(5)} \).

Step 3: Differentiate the inner function

The inner function is \( u(s) = 3^s - 2 \). The derivative of \( u(s) \) with respect to \( s \) is \( u'(s) = 3^s \ln(3) \).

Step 4: Apply the chain rule

Now, apply the chain rule by multiplying the derivative of the outer function by the derivative of the inner function. This gives:\( g'(s) = f'(u(s)) \cdot u'(s) = \frac{1}{(3^s - 2) \ln(5)} \cdot 3^s \ln(3) \).

Step 5: Simplify the expression

Combine and simplify the expression to find the derivative:\[ g'(s) = \frac{3^s \ln(3)}{(3^s - 2) \ln(5)} \] Thus, the derivative of \( g(s) \) with respect to \( s \) is \( \frac{3^s \ln(3)}{(3^s - 2) \ln(5)} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function changes. This is often referred to as finding the derivative. In more intuitive terms, if you have a curve that represents a function on a graph, differentiation helps you find the slope of that curve at any given point.

To perform differentiation, you need to follow certain established rules and understand the type of function you are dealing with:

• For a constant function such as a straight line, the derivative is 0 because the slope doesn't change.
• The derivative of a simple function like a power function is found using the power rule. This states that for a function in the form of \( x^n \), the derivative is \( nx^{n-1} \).
• If you're working with more complex functions, other rules like the chain rule or product rule might be needed.

Understanding differentiation is crucial because it is used to solve real-world problems, such as calculating speed, optimizing production, and predicting changing trends.

Chain Rule

The chain rule is a powerful tool for differentiation. It allows you to differentiate composite functions. These are functions inside other functions. The basic idea is to take the derivative of the outer function, keeping the inner function as it is, and then multiply this result by the derivative of the inner function.

Here's a clearer breakdown:

• Consider a composite function \( y = f(g(x)) \). This means you have \( y \) as a function \( f \) of some other function \( g \) of \( x \).
• The chain rule formula is: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This rule is particularly useful when differentiating logarithmic, exponential, and trigonometric functions that are composed with other functions.

By mastering the chain rule, you can tackle difficult calculus problems more easily. It is essential for dealing with complex real-world applications where simple rules aren't sufficient.

Logarithmic Differentiation

Logarithmic differentiation is a technique that simplifies the differentiation of complex functions. It's particularly useful when the function involves a variable in the base and the exponent. Here, you use the properties of logarithms to transform the problem into a simpler one.

To apply logarithmic differentiation:

• Take the natural logarithm (ln) of both sides of the function you want to differentiate.
• Use logarithmic identities to simplify, such as breaking down products and powers.
• Differentiate implicitly with respect to the independent variable.
• Finally, solve for the original derivative.

This method is incredibly helpful because logarithms turn multiplication into addition and power functions into products, which are easier to differentiate. It's especially handy for exponential functions, like \( 3^s \), which could be part of a larger expression, such as in the problem \( g(s) = \log_5(3^s - 2) \). Understanding logarithmic differentiation not only eases the process but also enhances your problem-solving toolkit.