Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Use the logarithmic differentiation to compute \(y^{\prime}(t)\) for a. \(y(t)=10^{t}\) b. \(y(t)=\frac{t-1}{t+1}\) c. \(y(t)=(t-1)^{3}\left(t^{3}-1\right)\) d. \(y(t)=(t-1)(t-2)(t-3)\) e. \(y(t)=u(t) v(t) w(t)\) f. \(y(t)=u(t) v(t)\)

Step-by-Step Solution

Answer

a: \(10^t \ln(10)\), b: \(\frac{2}{(t-1)(t+1)}\), c: \((t-1)^3(t^3-1)\left(\frac{3}{t-1} + \frac{3t^2}{t^3-1}\right)\), d: \((t-1)(t-2)(t-3)\left(\frac{1}{t-1} + \frac{1}{t-2} + \frac{1}{t-3}\right)\), e: \(uvw\left(\frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w}\right)\), f: \(uv\left(\frac{u'}{u} + \frac{v'}{v}\right)\)."

Step 1: Introduction to Logarithmic Differentiation

Logarithmic differentiation is a technique used primarily when dealing with products, quotients, or powers by applying the natural logarithm to each side of an equation. This simplifies differentiation, especially when dealing with complex expressions.

Step 1: Differentiate using logarithmic rules

Logarithms allow us to transform functions into sums and differences, making them easier to differentiate. We start by taking the natural logarithm of both sides of the equation and applying the chain rule during differentiation.

Step 2: Apply the property of logarithms to simplify

Use properties of logarithms such as \( \ln(a^b) = b \ln(a) \), \( \ln(ab) = \ln(a) + \ln(b) \), and \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \) to simplify the function before differentiating.

Step 4: Part a: Differentiate \( y(t) = 10^{t} \)

For \( y(t) = 10^t \): Start by taking the natural log of both sides: \( \ln(y) = \ln(10^t) \). Simplify using the property \( \ln(10^t) = t \ln(10) \). Differentiate both sides: \( \frac{1}{y} \cdot y' = \ln(10) \). Solve for \( y' \): \( y' = y \cdot \ln(10) = 10^t \cdot \ln(10) \).

Step 5: Part b: Differentiate \( y(t) = \frac{t-1}{t+1} \)

For \( y(t) = \frac{t-1}{t+1} \): Take the natural log: \( \ln(y) = \ln(t-1) - \ln(t+1) \). Differentiate: \( \frac{1}{y} y' = \frac{1}{t-1} - \frac{1}{t+1} \). Solve for \( y' \): Substitute \( y = \frac{t-1}{t+1} \) into the equation to get \( y' = \frac{y(((t+1) - (t-1))/(t-1)(t+1))} \). Simplifying, \( y' = \frac{2}{(t-1)(t+1)} \).

Step 6: Part c: Differentiate \( y(t) = (t-1)^3(t^3-1) \)

For \( y(t) = (t-1)^3(t^3-1) \): Taking the log of both sides, \( \ln(y) = 3 \ln(t-1) + \ln(t^3-1) \). Differentiate: \( \frac{1}{y}y' = \frac{3}{t-1} + \frac{3t^2}{t^3-1} \). Solve \( y' = y\left(\frac{3}{t-1} + \frac{3t^2}{t^3-1}\right) \) which means \( y' = (t-1)^3(t^3-1)\left(\frac{3}{t-1} + \frac{3t^2}{t^3-1}\right) \).

Step 7: Part d: Differentiate \( y(t) = (t-1)(t-2)(t-3) \)

For \( y(t) = (t-1)(t-2)(t-3) \): Taking log, \( \ln(y) = \ln(t-1) + \ln(t-2) + \ln(t-3) \). Differentiate: \( \frac{1}{y} y' = \frac{1}{t-1} + \frac{1}{t-2} + \frac{1}{t-3} \). Solve for \( y' = y\left(\frac{1}{t-1} + \frac{1}{t-2} + \frac{1}{t-3}\right) \) giving \( y' = (t-1)(t-2)(t-3)\left(\frac{1}{t-1} + \frac{1}{t-2} + \frac{1}{t-3}\right) \).

Step 8: Part e: Differentiate \( y(t) = u(t) v(t) w(t) \)

For \( y(t) = u(t) v(t) w(t) \), take log, \( \ln(y) = \ln(u) + \ln(v) + \ln(w) \). Differentiate: \( \frac{1}{y} y' = \frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w} \). Solve \( y' = y\left(\frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w}\right) \) leading to \( y' = u v w \left(\frac{u'}{u} + \frac{v'}{v} + \frac{w'}{w}\right) \).

Step 9: Part f: Differentiate \( y(t) = u(t) v(t) \)

For \( y(t) = u(t) v(t) \), log transform: \( \ln(y) = \ln(u) + \ln(v) \). Differentiate: \( \frac{1}{y} y' = \frac{u'}{u} + \frac{v'}{v} \). Solve \( y' = y\left(\frac{u'}{u} + \frac{v'}{v}\right) \) leading to \( y' = u v \left(\frac{u'}{u} + \frac{v'}{v}\right) \).

Key Concepts

Calculus

Calculus, in simple terms, is the branch of mathematics that studies continuous change. It is divided into two main parts: differential calculus and integral calculus.
Differential calculus concerns itself with the concept of a derivative, which measures how a function changes as its input changes. This is a fundamental tool used for understanding the behavior of functions.

• Calculus allows us to find areas under curves, slopes of curves, and numerous other useful insights into the behavior of functions.
• It provides the framework for solving problems related to motion and change in industrial, scientific, and mathematical settings.

Logarithmic differentiation, a technique derived from calculus, simplifies the differentiation of functions by leveraging the properties of logarithms. By transforming complex products, quotients, and powers into simpler sums and differences, we make calculations more manageable.

Chain Rule

The chain rule is a fundamental principle in calculus for finding the derivative of composite functions. It allows us to differentiate functions formed by the combination of two or more functions.
Imagine you have a function nested inside another. The chain rule enables you to differentiate this by taking the derivative of the outer function and then multiplying it by the derivative of the inner function.

• It is particularly useful in scenarios where functions are composed of multiple operations combined.
• The basic formula for the chain rule is \( (f(g(x)))' = f'(g(x)) \, g'(x) \).

In the context of logarithmic differentiation, the chain rule is often used after taking the natural logarithm of a function. This is crucial when dealing with the differentiation of powers and exponential terms, making the chain rule an essential tool for simplifying complex derivatives.

Derivatives

Derivatives are the cornerstone of differential calculus. They measure how a function changes as the input changes.
The derivative provides the rate at which a quantity changes and can be understood graphically as the slope of a tangent line to the curve at a point.

• The notation for a derivative with respect to a variable \( t \) is often represented as \( y'(t) \) or \( \frac{dy}{dt} \).
• Commonly used derivatives include those of polynomial, exponential, and trigonometric functions.

Derivatives are essential for optimization problems and in analyzing the motion of objects. In the case of logarithmic differentiation, derivatives are simplified by converting multiplication/division into addition/subtraction using logarithms before applying the chain or product rule.

Product Rule

The product rule is another key tool in calculus used to find the derivative of the product of two or more functions. This rule is particularly helpful when dealing with functions that are products of polynomial terms.
The product rule formula states that if you have two functions \( u(t) \) and \( v(t) \), then their derivative is given by:

• \( (uv)' = u'v + uv' \)

When using logarithmic differentiation, the product rule might be applied after converting a product into a sum using the properties of logarithms. The transformation helps manage the differentiation of each separate component individually before combining the results.
This makes the process more straightforward and reduces the likelihood of errors when solving complex equations.