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

Question

Compute \(P^{\prime}\) for: a. \(\quad P(t)=t^{2} e^{t}\) b. \(P(t)=\sqrt{t} e^{\sqrt{t}}\) c. \(\quad P(t)=\frac{t}{1+t^{2}}\) d. \(P(t)=\frac{t+1}{t-1}\) e. \(\quad P(t)=e^{t} \sqrt{1+t}\) f. \(\quad P(t)=t \ln t-t\) g. \(\quad P(t)=t e^{t}-e^{t}\) h. \(\quad P(t)=t^{2} e^{t}-2 t e^{t}+2 e^{t}\) i. \(\quad P(t)=\frac{\sqrt{t}}{\ln t}\) j. \(\quad P(t)=e^{t} \ln t\) k. \(P(t)=\frac{1}{\ln t}\) l. \(\quad P(t)=e^{(t \ln t)}\) m. \(P(t)=10 \frac{e^{0.2 t}}{9+e^{0.2 t}}\) n. \(\quad P(t)=\frac{20}{1+19 e^{-0.1 t}}\)

Step-by-Step Solution

Answer

Compute derivatives stepwise using product, chain, and quotient rules as needed.

Step 1: Differentiate Using the Product Rule

For part (a), the function is \(P(t) = t^2 e^t\). The product rule states that if you have two functions \(u(t) = t^2\) and \(v(t) = e^t\), then the derivative \(P'(t)\) is \(u'(t)v(t) + u(t)v'(t)\). Compute \(u'(t) = 2t\) and \(v'(t) = e^t\). Then, the derivative is \(P'(t) = 2t e^t + t^2 e^t = t(t+2)e^t\).

Step 2: Chain and Product Differentiation

For part (b), the function is \(P(t) = \sqrt{t} e^{\sqrt{t}}\). Let \(u(t) = \sqrt{t}\) and \(v(t) = e^{\sqrt{t}}\). Derivative \(u'(t) = \frac{1}{2\sqrt{t}}\) and for \(v(t)\), use the chain rule: \(v'(t) = e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}}\). The product rule then gives \(P'(t) = \frac{1}{2\sqrt{t}}e^{\sqrt{t}} + \sqrt{t} \cdot e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = e^{\sqrt{t}}(\frac{1}{2\sqrt{t}} + \frac{1}{2})\).

Step 3: Differentiate Using Quotient Rule

For part (c), \(P(t) = \frac{t}{1 + t^2}\). Using the quotient rule, if \(u(t) = t\) and \(v(t) = 1 + t^2\), \(P'(t) = \frac{(1)(1+t^2) - (t)(2t)}{(1+t^2)^2} = \frac{1+t^2-2t^2}{(1+t^2)^2} = \frac{1-t^2}{(1+t^2)^2}\).

Step 4: Derivative of Rational Functions

For part (d), \(P(t) = \frac{t+1}{t-1}\). Using the quotient rule, find \(P'(t) = \frac{(1)(t-1) - (t+1)(1)}{(t-1)^2} = \frac{(t-1)-(t+1)}{(t-1)^2}=\frac{-2}{(t-1)^2}\).

Step 5: Differentiate with Product and Chain Rule

For part (e), \(P(t) = e^t \sqrt{1+t}\). Treat \(u(t) = e^t\) and \(v(t) = \sqrt{1+t}\), with \(u'(t) = e^t\) and \(v'(t) = \frac{1}{2\sqrt{1+t}}\). Then, \(P'(t) = e^t\sqrt{1+t} + e^t \cdot \frac{1}{2\sqrt{1+t}}\), which simplifies to \(e^t(\sqrt{1+t} + \frac{1}{2\sqrt{1+t}})\).

Step 6: Logarithmic Differentiation

For part (f), \(P(t) = t\ln t - t\). Differentiate: \(P'(t) = 1\cdot\ln t + t\cdot\frac{1}{t} - 1 = \ln t + 1 - 1 = \ln t\).

Key Concepts

Product Rule

The Product Rule is a fundamental technique used in calculus to find the derivative of a product of two functions. If you have two functions, say \( u(t) \) and \( v(t) \), and you want to find the derivative of their product \( P(t) = u(t) v(t) \), you apply the product rule. The rule is articulated as:

• \[ P'(t) = u'(t) v(t) + u(t) v'(t) \]

This means you take the derivative of the first function \( u(t) \), multiply it by the second function \( v(t) \), and add it to the product of the first function with the derivative of the second. For example, if \( P(t) = t^2 e^t \), you break it down as \( u(t) = t^2 \) and \( v(t) = e^t \). Then \( u'(t) = 2t \) and \( v'(t) = e^t \), giving the derivative \( P'(t) = 2t e^t + t^2 e^t = t(t+2)e^t \). This versatility makes it highly useful for complex expressions where functions are multiplied.

Chain Rule

The Chain Rule is a crucial differentiation technique used when dealing with composite functions. If a function \( y = f(g(t)) \) is made up of two functions, such that \( y \) is a function of \( g(t) \) and \( g(t) \) is a function of \( t \), then the derivative \( y' \) with respect to \( t \) is found using:

• \[ y' = f'(g(t)) \,\ g'(t) \]

Simply put, you differentiate the outer function \( f \) with respect to \( g(t) \), leaving \( g(t) \) as it is, and then multiply by the derivative of the inner function \( g(t) \). Consider \( P(t) = \sqrt{t} e^{\sqrt{t}} \). We let the inner function \( u(t) = \sqrt{t} \) and the outer function \( v(t) = e^{u(t)} \). Here, the chain rule finds its role in differentiating \( e^{\sqrt{t}} \), resulting in the product rule application \( P'(t) = \frac{1}{2\sqrt{t}}e^{\sqrt{t}} + \sqrt{t} \cdot e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = e^{\sqrt{t}}(\frac{1}{2\sqrt{t}} + \frac{1}{2}) \).

Quotient Rule

The Quotient Rule aids in finding the derivative of a quotient of two functions. Given functions \( u(t) \) over \( v(t) \), the rule determines the derivative of the quotient \( P(t) = \frac{u(t)}{v(t)} \). The quotient rule is given by:

• \[ P'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{[v(t)]^2} \]

This involves taking the derivative of the top function \( u(t) \), multiplying by the bottom function \( v(t) \), and subtracting the product of the top function and derivative of the bottom function. Finally, the result is divided by the square of the bottom function \( v(t) \). For example, for \( P(t) = \frac{t}{1+t^2} \), identifying \( u(t) = t \) and \( v(t) = 1 + t^2 \), you apply the rule to find \( P'(t) = \frac{(1)(1+t^2) - (t)(2t)}{(1+t^2)^2} = \frac{1-t^2}{(1+t^2)^2} \). This formula is essential when functions appear in fractions.

Logarithmic Differentiation

Logarithmic Differentiation is a clever technique used for complex functions, especially useful when the function involves products or quotients raised to variable powers. The process involves taking the natural logarithm on both sides of the function \( y = f(x) \), differentiating, and then solving for \( y' \). This method reduces the complexity significantly by transforming multiplicative processes into additive ones via the logarithm.For instance, in solving \( P(t) = t \ln t - t \), we use the usual rules of differentiation without requiring a full application of logarithmic differentiation. By recognizing the form allows direct application of properties where necessary, you efficiently find \( P'(t) = \ln t + 1 - 1 = \ln t \). In more complex cases, taking the logarithm first can simplify handling variable exponents, particularly when multiple functions are intertwined.