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

Question

Compute the derivative of \(P\). a. \(P(t)=3 t^{2}-2 t+7\) b. \(P(t)=t+\frac{2}{t}\) c. \(\quad P(t)=\sqrt{t+2}\) d. \(P(t)=\left(t^{2}+1\right)^{5}\) e. \(\quad P(t)=\sqrt{2 t+1}\) f. \(\quad P(t)=2 t^{-3}-3 t^{-2}\) h. \(P(t)=(1+\sqrt{t})^{-1}\) g. \(P(t)=\frac{5}{t+5}\) i. \(P(t)=(1+2 t)^{5}\) j. \(P(t)=(1+3 t)^{1 / 3}\) k. \(P(t)=\frac{1}{1+\sqrt{t}}\) l. \(P(t)=\sqrt[3]{1+2 t}\)

Step-by-Step Solution

Answer

Derivatives are: (a) 6t-2, (b) 1-2t^{-2}, (c) \frac{1}{2\sqrt{t+2}}, (d) 10t(t^2+1)^4, (e) \frac{1}{\sqrt{2t+1}}, (f) -6t^{-4}+6t^{-3}, (h) -\frac{1}{2t^{1/2}(1+t^{1/2})^2}, (g) -\frac{5}{(t+5)^2}, (i) 10(1+2t)^4, (j) (1+3t)^{-2/3}, (k) -\frac{1}{2t^{1/2}(1+t^{1/2})^2}, (l) \frac{2}{3}(1+2t)^{-2/3}.

Step 1: Derivative of a Polynomial Function

For part (a), where \(P(t) = 3t^2 - 2t + 7\), we use the power rule for derivatives \(\frac{d}{dt}[t^n] = nt^{n-1}\). Consequently, the derivative is \(P'(t) = 6t - 2\).

Step 2: Derivative of a Fractional Polynomial

For part (b), the function is \(P(t) = t + \frac{2}{t}\). Rewrite it as \(t + 2t^{-1}\). Differentiating yields \(P'(t) = 1 - 2t^{-2}\).

Step 3: Derivative using Chain Rule for Square Root Function

Part (c) involves \(P(t) = \sqrt{t+2}\). Rewrite as \((t+2)^{1/2}\). Use the chain rule to find the derivative: \(P'(t) = \frac{1}{2}(t+2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{t+2}}\).

Step 4: Derivative of Composite Function using Chain Rule

For part (d), where \(P(t) = (t^2 + 1)^5\), use the chain rule again. Let \(u = t^2 + 1\), so \(\frac{du}{dt} = 2t\). The derivative is \(P'(t) = 5(u^4)\cdot \frac{du}{dt} = 10t(t^2 + 1)^4\).

Step 5: Chain Rule for Square Root Function

Part (e) has the function \(P(t) = \sqrt{2t+1}\). Rewrite as \((2t+1)^{1/2}\). Derivative is \(P'(t) = \frac{1}{2}(2t+1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2t+1}}\).

Step 6: Derivative of a Negative Exponent Function

In part (f), for \(P(t) = 2t^{-3} - 3t^{-2}\), differentiate each term separately: \(P'(t) = -6t^{-4} + 6t^{-3}\).

Step 7: Chain Rule for Inverse Function

For part (h), \(P(t) = (1 + \sqrt{t})^{-1}\). Rewrite as \((1+t^{1/2})^{-1}\). Use the chain rule: \(P'(t)=-1(1+t^{1/2})^{-2} \cdot \frac{1}{2}t^{-1/2} = -\frac{1}{2t^{1/2}(1+t^{1/2})^2}\).

Step 8: Quotient Rule Approach

In part (g), for \(P(t) = \frac{5}{t+5}\), use the quotient rule. Let \(f(t)=5\) and \(g(t)=t+5\), so \(P'(t) = \frac{g(t)f'(t) - f(t)g'(t)}{(g(t))^2} = \frac{0 \cdot (t+5) - 5 \cdot 1}{(t+5)^2} = \frac{-5}{(t+5)^2}\).

Step 9: Power Rule with Chain Rule

For part (i), where \(P(t) = (1+2t)^5\), apply the chain rule. Let \(u = 1+2t, \frac{du}{dt} = 2\). Then \(P'(t) = 5(1 + 2t)^4 \cdot 2 = 10(1+2t)^4\).

Step 10: Logarithmic Differentiation

Part (j) uses \(P(t) = (1+3t)^{1/3}\). Apply chain rule: let \(u = 1 + 3t\), giving \(P'(t) = \frac{1}{3}(1+3t)^{-2/3} \cdot 3 = (1+3t)^{-2/3}\).

Step 11: Chain Rule for Fraction Function

In part (k), for \(P(t) = \frac{1}{1+\sqrt{t}}\), rewrite as \((1 + t^{1/2})^{-1}\) and apply chain rule: \(P'(t) = -1(1 + t^{1/2})^{-2} \cdot \frac{1}{2}t^{-1/2} = -\frac{1}{2t^{1/2}(1+t^{1/2})^2}\).

Step 12: Derivative of a Cube Root Function

Lastly, for part (l), \(P(t) = \sqrt[3]{1+2t}\). Rewrite as \((1+2t)^{1/3}\). Use the chain rule: \(P'(t) = \frac{1}{3}(1+2t)^{-2/3} \cdot 2 = \frac{2}{3}(1+2t)^{-2/3}\).

Key Concepts

Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. The derivative of a function is often denoted by a prime symbol (\(f'(x)\)) or as \(\frac{df}{dx}\). To find the derivative, we use differentiation rules to explore how small changes in one variable affect another variable.
The process of finding a derivative is crucial in understanding the behavior of functions, like finding maximum and minimum points, and predicting future trends. Derivatives are used widely in various fields such as physics, engineering, and economics. Key operations involved in differentiation include using the power, chain, and quotient rules, each handling different forms of functions.

Power Rule

The power rule is a fundamental derivative rule used to find the derivative of any term in the form \(t^n\), where \(n\) is any real number. The rule states that if you have \(t^n\), the derivative is \(nt^{n-1}\). This means you multiply the power \(n\) by the term, then subtract one from the power.
For example, the derivative of \(3t^2\) is \(2 \times 3t^{2-1} = 6t\). The power rule is especially helpful in differentiating polynomial functions, where each term can be independently differentiated and then summed. It's straightforward and gives a quick way to handle many basic functions you’ll encounter in calculus.

Chain Rule

The chain rule is used to differentiate composite functions, which are functions within other functions. When you have a function like \( P(t) = (1 + 2t)^5 \), the chain rule helps you break it down into simpler parts.
You first differentiate the outer function while keeping the inner function the same, and then multiply by the derivative of the inner function. For example, if \( u = 1 + 2t \), finding the derivative \( \frac{d}{dt}(u^5) = 5u^4 \cdot \frac{du}{dt} = 10(1+2t)^4 \). The chain rule is essential in dealing with more complex expressions and ensures that every component of a function is accurately differentiated.

• First, identify the inner function and its derivative.
• Then, find the derivative of the outer function.
• Multiply them to obtain the derivative of the composite function.

Quotient Rule

The quotient rule is a technique used for finding the derivative of a function that is the division of two differentiable functions. It is particularly handy when the function is presented as a fraction. Let \(f(t)\) and \(g(t)\) be functions that are both differentiable. The quotient rule is stated as: \[ \frac{f}{g}(t) = \frac{g(t)f'(t) - f(t)g'(t)}{(g(t))^2} \]
This rule helps ensure that both the top and the bottom of the function are properly accounted for when differentiating. For example, when finding the derivative of \(P(t) = \frac{5}{t+5}\), apply the rule to find \(P'(t) = \frac{-5}{(t+5)^2}\). It’s crucial to handle the numerator and denominator separately and then combine them according to the formula, ensuring no part is ignored.