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

Question

Use the chain rule to differentiate \(P(t)\) for a. \(P(t)=e^{\left(-t^{2}\right)}\) b. \(P(t)=\left(e^{t}\right)^{2}\) c. \(P(t)=e^{2 \ln t}\) d. \(P(t)=\ln e^{2 t}\) e. \(P(t)=\ln (2 \sqrt{t})\) f. \(P(t)=\sqrt{2 \ln t}\) g. \(P(t)=\sqrt{e^{2 t}}\) h. \(P(t)=\sqrt{e^{\left(-t^{2}\right)}}\) i. \(\quad P(t)=\left(t+e^{-2 t}\right)^{4}\) j. \(P(t)=\left(1+e^{\left(-t^{2}\right)}\right)^{-1}\) k. \(P(t)=\frac{3}{4}\left(1-x^{2} / 16\right)^{1 / 2}\) 1\. \(P(t)=(t+\ln (1+2 t))^{2}\)

Step-by-Step Solution

Answer

Differentiate each function using the chain rule by identifying inner and outer functions, applying the rule, and simplifying results.

Step 1: Recognize the Function Composition

Identify the inner and outer functions for each part of the problem and prepare to apply the chain rule. For example, in part (a), the function can be seen as an exponential function where the exponent is an inner function of \(t\).

Step 2: Differentiate Using the Chain Rule

Apply the chain rule, which states that the derivative of a composite function \(f(g(t))\) is \(f'(g(t))g'(t)\). Here, differentiate the outer function first, and then multiply by the derivative of the inner function.

Step 3: Simplify the Derivative

Once the derivative is calculated, simplify it as much as possible to present the final result in its simplest form.

Step 4: Example Solution for One Part

For part (a), \(P(t)=e^{-t^2}\). The outer function is \(f(u) = e^u\) and the inner function is \(g(t) = -t^2\). Derivative of outer is \(e^u\), of the inner is \(-2t\). By chain rule: \(P'(t) = e^{-t^2}(-2t) = -2te^{-t^2}\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function is changing at any given point. When you differentiate a function, you obtain its derivative. The derivative represents the slope of the tangent line to the curve at a specific point. This is incredibly useful in many areas, from physics to economics.

To differentiate a function, you often apply derivative rules, such as the power rule, product rule, or chain rule. Each rule helps handle different types of functions. The power rule, for example, is used when you have a polynomial function. The chain rule, specifically, is very powerful when dealing with composite functions. It allows you to find the derivative of a nested function by breaking it down into its constituent parts.

Remember that mastery of differentiation requires practice. Your understanding will grow as you work through various examples and exercises, applying different rules as needed.

Composite Functions

Composite functions involve building a new function by combining two existing functions. These are often represented as \( f(g(x)) \), where \( g(x) \) is an inner function and \( f(u) \), where \( u = g(x) \), is the outer function.

In calculus, dealing with composite functions is central especially when differentiating them. The chain rule is the main tool used for this purpose.

Understanding composite functions requires recognizing the inner and outer parts clearly. Being able to distinguish between these helps in properly applying the chain rule. For example, in a problem like \(P(t) = e^{-t^2}\),the inner function is \(-t^2\) and the outer function is \(e^u\). A systematic approach starting by identifying these parts makes differentiation easier.

Practice identifying the components of composite functions in different contexts to improve your skills. This will allow you to tackle more complex calculus problems with confidence.

Calculus Education

In calculus education, the chain rule is a crucial component often introduced along with other differentiation techniques. It's essential for students to grasp this concept early, as it opens the door to understanding more complex functions.

An effective calculus education emphasizes the step-by-step approach to solving problems. This usually starts with identifying the components of the function and understanding the underlying principles before applying any specific differentiation methods.

Utilizing varied examples taken from real-world contexts can significantly enhance the learning experience. Seeing how calculus is applied outside the classroom not only aids comprehension but also demonstrates its practical importance.

To aid in understanding, it's beneficial to leverage visual aids and technology, like graphing software, to visualize functions and their derivatives. These resources can provide intuitive insights into how functions behave, reinforcing theoretical knowledge with a practical perspective.

Remember, studying calculus is not just about learning rules and formulas—it's about developing a way of thinking that can solve complex problems across diverse disciplines.