Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Use the logarithmic differentiation to compute \(y^{\prime}(t)\) for a. \(y(t)=\frac{(t-1)(t+1)}{t-2}\) b. \(y(t)=t e^{t}\) c. \(y(t)=e^{-\frac{t^{2}}{2}}\) d. \(y(t)=\sqrt{1+t^{2}}\) e. \(y(t)=\frac{t^{2}}{t^{2}+1}\) f. \(y(t)=2^{t}\) g. \(y(t)=b^{t} \quad b>0\) h. \(y(t)=\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}\) i. \(y(t)=\frac{\ln t}{e^{t}}\)

Step-by-Step Solution

Answer

Follow logarithmic differentiation steps for each function to find the derivative.

Step 1: Apply Logarithm to both sides for part a

Given \( y(t) = \frac{(t-1)(t+1)}{t-2} \). Take the natural logarithm of both sides: \( \ln y(t) = \ln\left(\frac{(t-1)(t+1)}{t-2}\right) \). Using log properties, this becomes: \( \ln y(t) = \ln(t-1) + \ln(t+1) - \ln(t-2) \).

Step 2: Differentiate Implicitly for part a

Differentiate both sides with respect to \( t \): \( \frac{1}{y(t)} y'(t) = \frac{1}{t-1} + \frac{1}{t+1} - \frac{1}{t-2} \). Multiply through by \( y(t) \) to solve for \( y'(t) \): \( y'(t) = y(t) \left( \frac{1}{t-1} + \frac{1}{t+1} - \frac{1}{t-2} \right) \). Substitute \( y(t) = \frac{(t-1)(t+1)}{t-2} \) into the equation to find \( y'(t) \).

Step 3: Repeat process for part b

For \( y(t) = t e^t \), take the logarithm: \( \ln y(t) = \ln t + t \). Differentiate: \( \frac{1}{y(t)} y'(t) = \frac{1}{t} + 1 \). Solve for \( y'(t) \): \( y'(t) = y(t) (\frac{1}{t} + 1) = t e^t \left( \frac{1}{t} + 1 \right) = e^t (1 + t) \).

Step 4: Repeat for part c

For \( y(t) = e^{-\frac{t^2}{2}} \), take the logarithm: \( \ln y(t) = -\frac{t^2}{2} \). Differentiate: \( \frac{1}{y(t)} y'(t) = -t \). Solve for \( y'(t) \): \( y'(t) = y(t) (-t) = -t e^{-\frac{t^2}{2}} \).

Step 5: Repeat for part d

For \( y(t) = \sqrt{1+t^2} \), take the logarithm: \( \ln y(t) = \frac{1}{2} \ln(1+t^2) \). Differentiate: \( \frac{1}{y(t)} y'(t) = \frac{1}{2} \cdot \frac{2t}{1+t^2} = \frac{t}{1+t^2} \). Solve for \( y'(t) \): \( y'(t) = \sqrt{1+t^2} \cdot \frac{t}{1+t^2} = \frac{t}{\sqrt{1+t^2}} \).

Step 6: Repeat for part e

For \( y(t) = \frac{t^2}{t^2+1} \), take the logarithm: \( \ln y(t) = \ln(t^2) - \ln(t^2+1) \). Differentiate: \( \frac{1}{y(t)} y'(t) = \frac{2t}{t^2} - \frac{2t}{t^2+1} \). Solve for \( y'(t) \): \( y'(t) = \frac{t^2}{t^2+1} \left( \frac{2t}{t^2} - \frac{2t}{t^2+1} \right) = \frac{2t^3}{(t^2+1)^2} \).

Step 7: Repeat for part f

For \( y(t) = 2^t \), take the logarithm: \( \ln y(t) = t \ln 2 \). Differentiate: \( \frac{1}{y(t)} y'(t) = \ln 2 \). Solve for \( y'(t) \): \( y'(t) = 2^t \ln 2 \).

Step 8: Repeat for part g

For \( y(t) = b^t \), take the logarithm: \( \ln y(t) = t \ln b \). Differentiate: \( \frac{1}{y(t)} y'(t) = \ln b \). Solve for \( y'(t) \): \( y'(t) = b^t \ln b \).

Step 9: Repeat for part h

For \( y(t) = \frac{e^t - e^{-t}}{e^t + e^{-t}} \), take the logarithm: \( \ln y(t) = \ln(e^t - e^{-t}) - \ln(e^t + e^{-t}) \). Differentiate: \( \frac{1}{y(t)} y'(t) = \frac{e^t + e^{-t}}{e^t - e^{-t}} - \frac{e^t - e^{-t}}{e^t + e^{-t}} \). Simplify to solve for \( y'(t) \).

Step 10: Repeat for part i

For \( y(t) = \frac{\ln t}{e^t} \), take the logarithm: \( \ln y(t) = \ln(\ln t) - t \). Differentiate: \( \frac{1}{y(t)} y'(t) = \frac{1}{t \ln t} - 1 \). Solve for \( y'(t) \): \( y'(t) = \frac{\ln t}{e^t} \left( \frac{1}{t \ln t} - 1 \right) \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful tool used in calculus to find the derivative of a function that is not easily expressed explicitly in terms of one variable. This technique is especially useful when dealing with equations where solving for one variable in terms of another is complex or perhaps impossible.
To perform implicit differentiation, follow these basic steps:

• Start by taking the derivative of both sides of the equation with respect to the variable you’re differentiating. Remember, if you're differentiating terms with the variable inside a function (like y), multiply by the derivative of that function (chain rule).
• Treat the dependent variable (often y) as a function of the independent variable (often x or t). As such, apply the chain rule when necessary.
• After differentiating, solve for the derivative of the dependent variable in terms of the independent variable.

In the context of logarithmic differentiation, the process is simplified since the natural log function can separate multiplicative terms and deal easily with quotients using log identities. This simplification allows the variable parts to be worked through implicitly and then tidied up algebraically at the end.

Calculus Techniques

Calculus is a branch of mathematics focusing on change and motion, and here, several techniques offer different ways to approach differentiation. These techniques are often applied in a coordinated fashion to solve more complex problems.
Here are some fundamental calculus techniques often utilized in differentiation:

• Product Rule: Used when differentiating an expression that involves the product of two functions. If you have two functions, u(t) and v(t), their derivative will be \( u'(t)v(t) + u(t)v'(t) \).
• Quotient Rule: Applies to expressions that are a ratio of two functions. For u(t)/v(t), the derivative is given by \( \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2} \).
• Chain Rule: Necessary for differentiating composite functions. If a function y = f(g(x)) is composed of two functions, the derivative is \( f'(g(x)) \cdot g'(x) \).

Combining these rules, we can solve a variety of challenging problems that appear in different forms. Utilizing logarithmic differentiation often involves applying these rules in tandem, especially the product and quotient rules, after the function is decomposed using logarithmic properties.

Differentiation of Exponential Functions

Differentiating exponential functions can sometimes seem daunting, but it becomes clearer with a solid understanding of the rules and properties related to exponential expressions.Here is a guide on how to handle differentiation for exponential functions:

• Natural Exponential Function (Base e): The derivative of the function e^x is simply e^x itself. More generally, for e^u(x), its derivative is given by e^u(x) times the derivative of u(x), based on the chain rule, i.e., \( e^{u(x)} \cdot u'(x) \).
• General Exponential Functions (Base b): When the base is not e, such as b^x, we use the natural logarithm, ln(b), in the differentiation: the derivative of b^x is \( b^{x} \ln(b) \). When differentiating an expression such as b^u(x), apply the chain rule, yielding \( b^{u(x)} \ln(b) \cdot u'(x) \).

Logarithmic differentiation simplifies the task considerably when dealing with exponential functions whose powers are themselves expressions. By taking the natural log of the function, we can break it into manageable parts, differentiate using straightforward rules, and subsequently resolve these into a neat solution. Logarithmic differentiation elegantly handles more complex scenarios involving exponential terms, often transforming multiplication into addition and division into subtraction due to logarithm properties, thus simplifying the overall process.