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Chapter 3: Problem 41

In Exercises \(1-56,\) find the derivatives. Assume that \(a\) and \(b\) are constants. $$w=\left(t^{2}+3 t\right)\left(1-e^{-2 t}\right)$$

Short Answer

Expert verified
The derivative of the function is \( w'(t) = 2t + 3 + (2t^2 + 4t - 3)e^{-2t} \).

Step by step solution

01

Identify the Function Type

In this exercise, we are given a function defined as a product of two sub-functions: \( w(t) = (t^2 + 3t)(1 - e^{-2t})\). This means we will need to apply the product rule to find the derivative. The product rule states that \((fg)' = f'g + fg'\) where \(f\) and \(g\) are functions of \(t\).
02

Define the Sub-functions

Let \( f(t) = t^2 + 3t \) and \( g(t) = 1 - e^{-2t} \). Our task is to find \( f'(t) \) and \( g'(t) \) in order to apply the product rule.
03

Differentiate the First Sub-function

Differentiate \( f(t) = t^2 + 3t \) with respect to \( t \). Using the power rule for derivatives, we find:\[ f'(t) = \frac{d}{dt}(t^2 + 3t) = 2t + 3 \]
04

Differentiate the Second Sub-function

Differentiate \( g(t) = 1 - e^{-2t} \) with respect to \( t \). Use the chain rule, where the derivative of \( e^{-2t} \) is \(-2e^{-2t}\):\[ g'(t) = \frac{d}{dt}(1 - e^{-2t}) = 0 + 2e^{-2t} = 2e^{-2t} \]
05

Apply the Product Rule

Now that we have both derivatives, we apply the product rule:\[ w'(t) = f'(t)g(t) + f(t)g'(t) \]Substitute \( f'(t) = 2t + 3 \), \( g(t) = 1 - e^{-2t} \), \( f(t) = t^2 + 3t \), and \( g'(t) = 2e^{-2t} \):\[ w'(t) = (2t + 3)(1 - e^{-2t}) + (t^2 + 3t)(2e^{-2t}) \]
06

Simplify the Derivative Expression

Expand each product:\[ w'(t) = (2t + 3) - (2t + 3)e^{-2t} + 2t^2e^{-2t} + 6te^{-2t} \]Combine like terms:\[ w'(t) = 2t + 3 - (2t + 3)e^{-2t} + 2t^2e^{-2t} + 6te^{-2t} \]
07

Final Simplification

Combine the common exponential terms:\[ w'(t) = 2t + 3 + (2t^2 + 6t - 2t - 3)e^{-2t} \]Simplify further:\[ w'(t) = 2t + 3 + (2t^2 + 4t - 3)e^{-2t} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
When differentiating functions that involve products of two or more sub-functions, the product rule is essential. It helps us determine the derivative of a function of the form \((fg)'\), where \(f\) and \(g\) are functions of the variable鈥攐ften \(t\).
According to the product rule, the derivative is calculated as:
  • \((fg)' = f'g + fg'\)
Here, \(f'\) and \(g'\) are the derivatives of \(f\) and \(g\) respectively.
For instance, in a problem where the function is \(w(t) = (t^2 + 3t)(1 - e^{-2t})\), define \(f(t) = t^2 + 3t\) and \(g(t) = 1 - e^{-2t}\).
  • Find each derivative \(f'(t)\) and \(g'(t)\) through methods like the chain or power rule.
  • Plug these values back into the product rule formula to find \(w'(t)\).
This method systematically breaks down complex expressions by mitigating errors and confusion during the differentiation process.
Chain Rule
The chain rule is a critical tool in calculus for finding derivatives of composite functions. It is especially useful when dealing with nested functions, like \(e^{-2t}\) in our example.
The chain rule states:
  • If you have a composite function \(c(x) = g(h(x))\), then the derivative \(c'(x) = g'(h(x)) \cdot h'(x)\).
To apply this to our case, consider how you differentiate \(e^{-2t}\):
  • If \(h(t) = -2t\), then \(g(u) = e^u\).
  • Differentiate \(h(t)\) to get \(h'(t) = -2\).
  • Since \(g(u) = e^u\), \(g'(u) = e^u\).
  • Thus, \(\frac{d}{dt}(e^{-2t}) = e^{-2t} \cdot (-2) = -2e^{-2t}\).
Understanding the chain rule is crucial for functions where derivatives involve inner functions or when exponents are not constant.
Power Rule
The power rule is one of the simplest and most commonly used rules for differentiation, particularly useful for polynomial functions. When you have a term of the form \(t^n\), the power rule states that the derivative is:
  • \(\frac{d}{dt}(t^n) = n \cdot t^{n-1}\)
For example, if \(f(t) = t^2 + 3t\):
  • Apply the power rule to each term: \(\frac{d}{dt}(t^2) = 2t^{2-1} = 2t\) and \(\frac{d}{dt}(3t) = 3\cdot 1 \cdot t^{1-1} = 3\).
Hence, \(f'(t) = 2t + 3\).
This rule greatly simplifies finding derivatives since it reduces computation to straightforward arithmetic operations. In combination with other rules like the product or chain rule, it forms a cornerstone of calculus study.

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Most popular questions from this chapter

Are the statements true or false for a function \(f\) whose domain is all real numbers? If a statement is true, explain how you know. If a statement is false, give a counterexample. If \(f^{\prime}(x) \geq 0\) for all \(x,\) then \(f(a) \leq f(b)\) whenever \(a \leq b\)

Simplify the expressions. $$\cosh (\ln t)$$

Explain what is wrong with the statement. If \(w(x)=\ln \left(1+x^{4}\right)\) then \(w^{\prime}(x)=1 /\left(1+x^{4}\right).\)

Use the table and the fact that \(f(x)\) is invertible and differentiable everywhere to find \(\left(f^{-1}\right)^{\prime}(3).\) $$\begin{array}{c|c|c} \hline x & f(x) & f^{\prime}(x) \\ \hline 3 & 1 & 7 \\ \hline 6 & 2 & 10 \\\ \hline 9 & 3 & 5 \\ \hline \end{array}$$

Explain what is wrong with the statement. The derivative of \(f(x)=\ln (\ln x)\) is $$f^{\prime}(x)=\frac{1}{x} \ln x+\ln x \frac{1}{x}=\frac{2 \ln x}{x}$$.
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