Problem 15 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A useful special case of the quotient formula is the reciprocal formula: If \(u(t)\) has a derivative and \(u(t) \neq 0\) and $$ P(t)=\frac{1}{u(t)} $$ then $$ P^{\prime}(t)=\frac{-1}{u^{2}(t)} u^{\prime}(t) $$ Prove the formula using logarithmic differentiation. That is, write $$ \ln P(t)=\ln \left(\frac{1}{u(t)}\right)=-\ln u(t) $$ and compute the derivatives of both sides using the logarithm chain rule. We write the formula as $$ \left[\frac{1}{u(t)}\right]^{\prime}=\frac{-1}{u^{2}(t)} u^{\prime}(t) \quad \text { Reciprocal Rule } $$

Step-by-Step Solution

Answer

Differentiate \( \ln P(t) = -\ln u(t) \) to find \( P'(t) = \frac{-1}{u^2(t)} u'(t) \).

Step 1: Express the Function Logarithmically

Given \( P(t) = \frac{1}{u(t)} \), take the natural logarithm of both sides. Write \( \ln P(t) = \ln \left( \frac{1}{u(t)} \right) \). Since \( \ln \left( \frac{1}{x} \right) = -\ln x \), we have \( \ln P(t) = -\ln u(t) \).

Step 2: Differentiate Using the Logarithmic Chain Rule

Differentiate both sides with respect to \( t \). On the left side, differentiate \( \ln P(t) \) to get \( \frac{1}{P(t)} P'(t) \) by applying the chain rule. On the right side, differentiate \( -\ln u(t) \) to get \( -\frac{u'(t)}{u(t)} \) using the chain rule for logarithms.

Step 3: Relate the Derivatives

From Step 2, we have \( \frac{1}{P(t)} P'(t) = -\frac{u'(t)}{u(t)} \). Substitute \( P(t) = \frac{1}{u(t)} \) to transform \( \frac{1}{P(t)} \) into \( u(t) \). So, \( u(t) P'(t) = -\frac{u'(t)}{u(t)} \).

Step 4: Solve for \( P'(t) \)

Multiplying both sides of the equation by \( u(t) \), we isolate \( P'(t) \): \( P'(t) = -\frac{u'(t)}{u^2(t)} \). This confirms the Reciprocal Rule formula \( P'(t) = \frac{-1}{u^2(t)} u'(t) \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a clever technique used to differentiate functions that are complicated products or quotients.
To apply this method, we take the natural logarithm of the function, which simplifies the differentiation process, especially for functions that involve exponentials or products.
In this particular exercise, we use logarithmic differentiation to prove the derivative of a reciprocal function. By rewriting the function in logarithmic form, the differentiation becomes more manageable.

• Convert the function using logarithms.
• Differentiate both sides using the rules of logarithms.
• Use the results to find the derivative of the original function.

This method is especially handy when dealing with functions like \(P(t) = \frac{1}{u(t)}\), which can be transformed by taking the natural logarithm, \(\ln P(t) = -\ln u(t)\). This transformation simplifies the differentiation process significantly which in turn helps us neatly derive the reciprocal rule.

Derivative of Reciprocal Function

The derivative of a reciprocal function holds a unique formula due to its specific structure.
When you have a function \(P(t) = \frac{1}{u(t)}\), recognizing its reciprocal nature allows us to apply a standard method to find its derivative.
The reciprocal rule formula is given by: \[P'(t) = \frac{-1}{u^2(t)} u'(t)\]

• Understand that the derivative starts with a negative sign since a complex function, \(\frac{1}{x}\), naturally results in \(-\frac{1}{x^2}\).
• The derivative of the inside function, \(u(t)\), multiplies the entire expression, due to the Chain Rule.

This formula essentially breaks down into the negative reciprocal of the square of the function, multiplied by its derivative. This consistent pattern emerges thanks to the properties of the reciprocal function and plays a crucial role in calculus, especially in physics and engineering, where such functions frequently occur.

Quotient Rule

The Quotient Rule is a fundamental differentiation rule used when dealing with the division of two functions.
It states how to obtain the derivative of a quotient \(\frac{v(t)}{u(t)}\): \[\left( \frac{v(t)}{u(t)} \right)^{\prime} = \frac{v^{\prime}(t)u(t) - v(t)u^{\prime}(t)}{[u(t)]^2}\]In the context of the reciprocal function, while we didn't directly apply the quotient rule in the initial solution process, it's important to see the connection.

• The reciprocal of a function \(u(t)\) can be viewed through the lens of the quotient rule as \(\frac{1}{u(t)}\), embodying the principle of division.
• The formula derived in our problem from logarithmic differentiation aligns with what results from the quotient rule.

By understanding the quotient rule, you can appreciate the consistency across different derivative rules, creating a harmonious framework within calculus.

Chain Rule in Differentiation

The Chain Rule is crucial in calculus for finding the derivative of composed functions.
When you have a function inside another function, like \(g(h(t))\), the derivative becomes the product of the derivatives of the outer function and the inner function.
The rule states:\[[g(h(t))]' = g'(h(t)) \cdot h'(t)\]In our exercise, the chain rule was used during logarithmic differentiation:

• Applying the chain rule to the logarithmic representation \(\ln P(t) = -\ln u(t)\) involves differentiating \(\ln P(t)\), meaning \(\frac{1}{P(t)} P'(t)\).
• The chain rule also translates to the differentiation of \(-\ln u(t)\), yielding \(-\frac{u'(t)}{u(t)}\).

This methodology highlights how the chain rule simplifies more involved derivatives, ensuring accurate and systematic differentiation even in complicated compositions. By familiarizing with the chain rule, understanding differentiation across various contexts becomes far easier.