Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

The word differentiate means 'find the derivative of'. Differentiate a. \(P(t)=\frac{e^{-3 t}}{t^{2}}\) d. \(P(t)=t^{2} e^{2 t}\) g. \(P(t)=\left(e^{t}\right)^{5}\) b. \(P(t)=e^{2 \ln t}\) e. \(P(t)=e^{t \ln 2}\) h. \(P(t)=\frac{t-1}{t+1}\) c. \(P(t)=e^{t} \ln t\) f. \(P(t)=e^{2}\) i. \(P(t)=\frac{3 t^{2}-2 t-1}{t^{-} 1}\)

Step-by-Step Solution

Answer

Solutions involve applying quotient rule, product rule, and simplifying exponents; derivative of a constant is zero.

Step 1: Differentiate using the Quotient Rule

To differentiate \( P(t) = \frac{e^{-3t}}{t^2} \), apply the quotient rule, which states that if you have a function \( \frac{u}{v} \), its derivative is \( \frac{u'v - uv'}{v^2} \). Here, \( u = e^{-3t} \) and \( v = t^2 \). Start by differentiating both: \( u' = -3e^{-3t} \) and \( v' = 2t \). Then, substitute into the formula:\[ P'(t) = \frac{(-3 e^{-3t}) t^2 - (e^{-3t})(2t)}{(t^2)^2} \].

Step 2: Simplify the Derivative

Simplify the expression from the quotient rule. Combine and simplify terms:\[ P'(t) = \frac{-3 e^{-3t} t^2 - 2t e^{-3t}}{t^4} = \frac{-e^{-3t} (3t^2 + 2t)}{t^4} \].

Step 3: Differentiate using the Product Rule

For \( P(t) = t^2 e^{2t} \), use the product rule, which states that if you have a function \( uv \), its derivative is \( u'v + uv' \). Here, \( u = t^2 \), \( v = e^{2t} \). Then, \( u' = 2t \), \( v' = 2e^{2t} \). Compute the derivative:\[ P'(t) = 2t e^{2t} + t^2 (2e^{2t}) = 2te^{2t} + 2t^2 e^{2t} \].

Step 4: Simplify using Exponent Rules

For \( P(t) = (e^t)^5 \), rewrite the expression using exponent rules: \( e^{5t} \). Then differentiate: \( P'(t) = 5e^{5t} \).

Step 5: Derivative using Exponent and Logarithm Properties

For \( P(t) = e^{2 \, \ln t} \), recognize \( e^{2 \, \ln t} = t^2 \). Differentiate \( t^2 \) to obtain: \( P'(t) = 2t \).

Step 6: Exponential Derivative

Given \( P(t) = e^{t \, \ln 2} \), simplify to get: \( e^{\ln(2^t)} = 2^t \). Derivative is \( P'(t) = 2^t \ln 2 \).

Step 7: Derivative using Product and Logarithmic Differentiation

For \( P(t) = e^t \ln t \), apply the product rule: \((u'v + uv')\). Here, \( u = e^t \), \( v = \ln t \). Then \( u' = e^t \), \( v' = \frac{1}{t} \). Compute:\[ P'(t) = e^t \ln t + e^t \frac{1}{t} = e^t \left( \ln t + \frac{1}{t} \right) \].

Step 8: Constant Differentiation

For \( P(t) = e^2 \), which is a constant, the derivative is zero, as the derivative of any constant is zero.

Step 9: Quotient Rule for Rational Function

For \( P(t) = \frac{3t^2 - 2t - 1}{t - 1} \), apply the quotient rule again:\[ \frac{u}{v} \rightarrow \frac{u'v - uv'}{v^2} \], where \( u = 3t^2 - 2t - 1 \), \( v = t - 1 \). Find \( u' = 6t - 2 \) and \( v' = 1 \), leading to:\[ P'(t) = \frac{(6t - 2)(t - 1) - (3t^2 - 2t - 1)(1)}{(t-1)^2} \]. Simplify and solve.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental tool in calculus used for differentiating functions that are expressed as a ratio of two functions, specifically a numerator divided by a denominator. It helps us find the derivative of a function that can be written in the form \( \frac{u}{v} \). The rule states that the derivative of such a function is given by the formula:

• \( \frac{u'v - uv'}{v^2} \)

Here, \( u \) and \( v \) are functions of \( t \), and \( u' \) and \( v' \) are their derivatives respectively.

To better understand this, consider the example \( P(t) = \frac{e^{-3t}}{t^2} \). Apply the quotient rule by first differentiating both the numerator \( u = e^{-3t} \) and the denominator \( v = t^2 \) separately. With \( u' = -3e^{-3t} \) and \( v' = 2t \), we insert these into the formula to get:

• \( P'(t) = \frac{(-3 e^{-3t}) t^2 - (e^{-3t})(2t)}{t^4} \)

By simplifying, we combine the terms in the numerator and obtain the derivative. This technique is particularly useful for more complex rational functions.

Product Rule

The Product Rule is an essential differentiation rule used when two functions are multiplied together. When you have a function \( P(t) = u(t)v(t) \), the derivative is calculated as:

• \( u'v + uv' \)

This means you differentiate each function separately, then add their products.

Let's see this in practice with \( P(t) = t^2 e^{2t} \). Identify \( u = t^2 \) and \( v = e^{2t} \). Next, find their derivatives: \( u' = 2t \) and \( v' = 2e^{2t} \). Plug these derivatives into the product rule formula:

• \( P'(t) = 2t e^{2t} + t^2 (2e^{2t}) \)

The resulting expression shows the rate of change of the function \( P(t) \). The product rule is useful for expressions where direct differentiation is not straightforward and caters to various combinations of functions.

Exponential Functions

Exponential functions are crucial in calculus due to their unique properties and frequent occurrence in mathematical models, particularly in growth and decay problems. An exponential function can generally be represented as\( e^{kx} \), where \( k \) is a constant. The derivative of an exponential function \( e^{kx} \) is simply \( ke^{kx} \), reflecting how it retains its form under differentiation.

Consider \( P(t) = (e^t)^5 \). This can be rewritten using exponent laws as \( e^{5t} \).

• Differentiate \( e^{5t} \) to yield \( P'(t) = 5e^{5t} \).

This derivative calculation demonstrates how exponential functions simplify the differentiation process and why they are often used in models requiring constant growth rates.

Logarithmic Differentiation

Logarithmic Differentiation is a smart technique to simplify complex functions, particularly those involving products, quotients, or powers that can be bothersome to differentiate directly. It leverages the properties of logarithms to make the differentiation process more manageable.

For instance, consider \( P(t) = e^{2 \ln t} \). Recognize that \( e^{2 \ln t} \) simplifies to \( t^2 \). Therefore, differentiating \( t^2 \) directly gives \( P'(t) = 2t \). This method is particularly helpful in dealing with functions that involve exponentiation and logarithmic expressions by converting them into a more straightforward form.