Problem 13 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(P(t)=u(t) \times v(t) .\) Then $$ \ln P(t)=\ln (u(t) \times v(t))=\ln u(t)+\ln v(t) $$ Compute the derivative of the two sides of Equation 6.11 using the logarithm chain rule and show that $$ P^{\prime}(t)=u(t) v^{\prime}(t)+u^{\prime}(t) v(t) $$

Step-by-Step Solution

Answer

\( P'(t) = u(t)v'(t) + u'(t)v(t) \) is derived by differentiating and using the chain rule.

Step 1: Express the Problem

The problem involves taking the natural logarithm of a product function and then computing the derivative of both sides. We start with \( \ln P(t) = \ln (u(t) \times v(t)) = \ln u(t) + \ln v(t) \).

Step 2: Apply the Logarithmic Property

Using the property of logarithms, \( \ln(a \times b) = \ln a + \ln b \), we express the logarithm of the product as a sum of logarithms: \( \ln P(t) = \ln u(t) + \ln v(t) \).

Step 3: Differentiate Both Sides

Differentiate each side of \( \ln P(t) = \ln u(t) + \ln v(t) \) with respect to \( t \).

Step 4: Differentiate the Left Side Using the Chain Rule

The derivative of \( \ln P(t) \) is \( \frac{1}{P(t)} \times P'(t) \) by the chain rule, because the derivative of \( \ln x \) is \( \frac{1}{x} \).

Step 5: Differentiate the Right Side Using the Chain Rule

The derivative of \( \ln u(t) + \ln v(t) \) is \( \frac{1}{u(t)} \times u'(t) + \frac{1}{v(t)} \times v'(t) \) using the chain rule.

Step 6: Set the Derivatives Equal

Equating the derivatives, we have: \[ \frac{1}{P(t)} P'(t) = \frac{1}{u(t)} u'(t) + \frac{1}{v(t)} v'(t) \]

Step 7: Solve for \( P'(t) \)

Multiply through by \( P(t) = u(t) \times v(t) \) to clear the fraction, leading to: \[ P'(t) = u(t) \times v(t) \left( \frac{1}{u(t)} u'(t) + \frac{1}{v(t)} v'(t) \right) \]Simplifying, \[ P'(t) = v(t)u'(t) + u(t)v'(t) \]

Step 8: Conclusion

We confirmed that \( P'(t) = u(t)v'(t) + u'(t)v(t) \). This demonstrates that the derivative of the product function \( P(t) \) follows this expression.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. When you have a function nested within another function, the chain rule is applied. It states that to differentiate a composite function \( f(g(x)) \), you compute the derivative of the outer function \( f \) with respect to its inner function \( g \), and then multiply by the derivative of the inner function \( g \) with respect to the variable \( x \).

This is mathematically expressed as:

• \( (f(g(x)))' = f'(g(x)) \, g'(x) \)

In the context of logarithmic differentiation, when we take the derivative of \( \ln P(t) \), it is seen as taking the derivative of the natural logarithm of \( P(t) \), which involves the chain rule.

For \( \ln P(t) \), the derivative using the chain rule is \( \frac{1}{P(t)} \times P'(t) \), because we first differentiate the natural logarithm and then multiply it by the derivative of the inner function \( P(t) \).

Product Rule

The product rule is crucial when dealing with derivatives of products of functions. It allows you to differentiate expressions where two or more functions are multiplied together.

The rule states that for functions \( u(t) \) and \( v(t) \), the derivative of their product is:

• \( (u(t) \, v(t))' = u(t) \, v'(t) + u'(t) \, v(t) \)

This means you take the derivative of the first function \( u(t) \) while leaving the second function \( v(t) \) unchanged, plus the derivative of the second function \( v(t) \) while leaving the first function \( u(t) \) unchanged.

In the final steps of our problem, you can see the rearrangement of terms using the product rule applied after using logarithmic differentiation. It demonstrates that each part contributes additively to the final derivative \( P'(t) \).

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational and transcendental constant approximately equal to 2.71828. The natural logarithm is unique due to its mathematical properties and relationships in calculus.

In this exercise, the natural logarithm is used to simplify the differentiation process by converting the product of functions into a sum. This is because \( \ln(u(t) \, v(t)) \) can be expanded as \( \ln u(t) + \ln v(t) \), making it easier to differentiate.

The derivative of \( \ln x \) is particularly straightforward:

• \( \frac{d}{dx} (\ln x) = \frac{1}{x} \)

When using the natural logarithm in logarithmic differentiation, it simplifies many complex expressions, enabling the application of other key rules like the chain rule or product rule more effectively.

Derivative

A derivative is a tool in calculus that measures how a function changes as its input changes. It is the foundational concept for understanding changes and motion in mathematics and physics. The derivative of a function at a specific point tells us the rate at which the function is changing at that point.

The concept of the derivative is visually seen as the slope of the tangent line to the curve of a function at a given point. Mathematically, it is defined as:

• \( f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \)

In our specific exercise, derivatives are used to find the rate of change of \( P(t) \) as derived from the functions \( u(t) \) and \( v(t) \). By finding \( P'(t) \), we determine how the product function changes, using techniques like the chain rule and product rule, allowing us to carefully dissect and solve different parts of the derivative problem.