Problem 1
Question
Use one rule for each step and identify the rule to differentiate a. \(P(t)=3 \ln t+e^{3 t}\) b. \(P(t)=t^{2}+\ln 2 t\) c. \(P(t)=\ln 5\) d. \(P(t)=\ln \left(e^{2 t}\right)\) e. \(P(t)=\ln \left(t^{2}+t\right)\) f. \(P(t)=e^{t^{2}-t}\) g. \(P(t)=e^{1 / x}\) h. \(P(t)=e^{\sqrt{x}}\) i. \(P(t)=\ln \left((t+1)^{2}\right)\) j. \(P(t)=e^{-t^{2} / 2}\)
Step-by-Step Solution
Verified Answer
Differentiate each function using appropriate differentiation rules, applying simplifications as required.
1Step 1: Differentiate part (a) using sum, constant and exponential rules
To differentiate \(P(t)=3 \ln t+e^{3 t}\), we take the derivative of each term separately. The derivative of \(3 \ln t\) is \(\frac{3}{t}\) using the constant multiple rule and the derivative of \(\ln t\), which is \(\frac{1}{t}\). The derivative of \(e^{3t}\) is \(3e^{3t}\) using the chain rule. So, \(P'(t) = \frac{3}{t} + 3e^{3t}\).
2Step 2: Differentiate part (b) using power and product rules
For \(P(t)=t^{2} + \ln 2t\), the derivative of \(t^2\) is \(2t\) by the power rule. For \(\ln 2t\), use the chain and product rules: \(\frac{d}{dt}(\ln 2t) = \frac{1}{2t} \cdot 2 = \frac{1}{t}\). Thus, \(P'(t) = 2t + \frac{1}{t}\).
3Step 3: Differentiate part (c) as a constant function
Since \(P(t)=\ln 5\) is a constant, its derivative is 0. The rule applied here is the constant rule, where the derivative of any constant is zero. Therefore, \(P'(t) = 0\).
4Step 4: Differentiate part (d) using logarithm rules
For \(P(t)=\ln(e^{2t})\), simplify first: \(\ln(e^{2t}) = 2t\) using the property \(\ln(e^{a})=a\). The derivative is then \(2\). Therefore, \(P'(t) = 2\).
5Step 5: Differentiate part (e) using the chain rule with logarithm
Given \(P(t)=\ln(t^{2}+t)\), apply the chain rule: \(\frac{1}{(t^2+t)} \cdot (2t+1)\). Thus, \(P'(t) = \frac{2t+1}{t^2+t}\).
6Step 6: Differentiate part (f) using the chain rule and exponential functions
For \(P(t)=e^{t^2-t}\), apply the chain rule: \(e^{t^2-t} \cdot (2t-1)\). Therefore, \(P'(t) = (2t-1)e^{t^2-t}\).
7Step 7: Differentiate part (g) using substitution and chain rule
With \(P(t)=e^{1/t}\), apply chain rule: \(-\frac{1}{t^2} e^{1/t}\). Thus, \(P'(t) = -\frac{1}{t^2} e^{1/t}\).
8Step 8: Differentiate part (h) using chain rule for square root
For \(P(t)=e^{\sqrt{x}}\), chain rule gives: \(\frac{1}{2\sqrt{x}}e^{\sqrt{x}}\). Thus, \(P'(t) = \frac{1}{2\sqrt{x}}e^{\sqrt{x}}\).
9Step 9: Differentiate part (i) using chain and power rules
For \(P(t)=\ln((t+1)^2)\), simplify first: \(2 \ln(t+1)\). The derivative is \(\frac{2}{t+1}\). Thus, \(P'(t)=\frac{2}{t+1}\).
10Step 10: Differentiate part (j) using chain rule on exponential
For \(P(t)=e^{-t^2/2}\), apply the chain rule: \(-t \cdot e^{-t^2/2}\). Therefore, \(P'(t)= -t \cdot e^{-t^2/2}\).
Key Concepts
Derivative rulesChain ruleLogarithmic differentiationExponential functions
Derivative rules
To find the derivative of a function, various rules and techniques are essential. Here are some fundamental derivative rules used frequently in calculus exercises such as this one.
- Constant Rule: This rule states that the derivative of a constant is 0. For example, if \(P(t) = \ln 5\), then \(P'(t) = 0\) because \(5\) is a constant.
- Power Rule: This rule is used for polynomials. To find the derivative of \(t^n\), bring down the exponent and multiply it by the power of \(t\) reduced by one. For example, the derivative of \(t^2\) is \(2t\).
- Sum Rule: The derivative of a sum of functions is the sum of their derivatives. For instance, if \(P(t) = 3 \ln t + e^{3t}\), differentiate each term separately.
- Exponential Functions Rule: The derivative of \(e^u\) (where \(u\) is a function of \(t\)) is \(e^u \cdot u'\). This is used extensively when differentiating exponential functions.
Chain rule
The chain rule is a powerful tool in calculus for finding the derivative of a composite function. It's essential when functions are nested within others.
Apply the chain rule when you see a function inside another function. Mathematically, if \(y = f(g(t))\), the chain rule tells you to multiply the derivative of the outer function \(f(g(t))\) by the derivative of the inner function \(g(t)\).
In the exercise, the chain rule was used multiple times, such as in part \(f\), where \(P(t) = e^{t^2-t}\). Here, the derivative was \(e^{t^2-t} \cdot (2t - 1)\). The outer function is \(e^x\), while the inner function is \(t^2 - t\). First, differentiate \(e\) to itself, and then multiply it by the derivative of \(t^2-t\).
Another clear application is in part \(g\) with \(P(t) = e^{1/t}\). Here, the derivative is determined by first tackling the outer function \(e\) and then the inner fraction \(\frac{1}{t}\).
Apply the chain rule when you see a function inside another function. Mathematically, if \(y = f(g(t))\), the chain rule tells you to multiply the derivative of the outer function \(f(g(t))\) by the derivative of the inner function \(g(t)\).
In the exercise, the chain rule was used multiple times, such as in part \(f\), where \(P(t) = e^{t^2-t}\). Here, the derivative was \(e^{t^2-t} \cdot (2t - 1)\). The outer function is \(e^x\), while the inner function is \(t^2 - t\). First, differentiate \(e\) to itself, and then multiply it by the derivative of \(t^2-t\).
Another clear application is in part \(g\) with \(P(t) = e^{1/t}\). Here, the derivative is determined by first tackling the outer function \(e\) and then the inner fraction \(\frac{1}{t}\).
Logarithmic differentiation
Logarithmic differentiation is a method that involves using properties of logarithms to simplify the differentiation process. This is particularly useful when dealing with products, quotients, or powers of functions that are otherwise cumbersome to differentiate.
This technique shines especially with natural logs, \(\ln\), since the derivative of \(\ln u\) is \(\frac{1}{u} \cdot u'\). You see this in action in part \(e\) where \(P(t)=\ln(t^2+t)\). The derivative required you to first simplify \(\ln(t^2+t)\) into its constituents. Using the chain rule and \(\ln\)'s properties, you obtain \(\frac{2t+1}{t^2+t}\).
Logarithmic differentiation is also useful when functions have exponents that aren't constants. For instance, when differentiating \(\ln((t+1)^2)\) in part \(i\), the function logs both the base and exponent, simplifying the process and leading directly to a derivative of \(\frac{2}{t+1}\).
This technique shines especially with natural logs, \(\ln\), since the derivative of \(\ln u\) is \(\frac{1}{u} \cdot u'\). You see this in action in part \(e\) where \(P(t)=\ln(t^2+t)\). The derivative required you to first simplify \(\ln(t^2+t)\) into its constituents. Using the chain rule and \(\ln\)'s properties, you obtain \(\frac{2t+1}{t^2+t}\).
Logarithmic differentiation is also useful when functions have exponents that aren't constants. For instance, when differentiating \(\ln((t+1)^2)\) in part \(i\), the function logs both the base and exponent, simplifying the process and leading directly to a derivative of \(\frac{2}{t+1}\).
Exponential functions
Exponential functions are those of the form \(e^u\) where \(e\) is the base of the natural logarithm, and \(u\) is a function of \(t\). Differentiating these involves knowledge of both basic exponential properties and the chain rule.
For an exponential function, the derivative of \(e^u\) is \(e^u \cdot \frac{du}{dt}\). This pattern repeats throughout the exercise, such as with \(e^{3t}\), \(e^{t^2-t}\), and \(e^{-t^2/2}\).
Understanding these patterns greatly assists in differentiating more complex exponential functions effortlessly.
For an exponential function, the derivative of \(e^u\) is \(e^u \cdot \frac{du}{dt}\). This pattern repeats throughout the exercise, such as with \(e^{3t}\), \(e^{t^2-t}\), and \(e^{-t^2/2}\).
- For \(e^{3t}\), differentiate simply by multiplying the unchanged base \(e^{3t}\) by 3, the derivative of the exponent.
- For \(e^{t^2-t}\), use the chain rule. Multiply the initial function by the derivative of the inner expression \(t^2-t\), which brings out \(2t-1\).
- Handling \(e^{-t^2/2}\) means tackling the exponent's derivative separately, resulting in \(-t\).
Understanding these patterns greatly assists in differentiating more complex exponential functions effortlessly.
Other exercises in this chapter
Problem 1
Compute \(P^{\prime}(t)\) for: a. \(P(t)=e^{5 t}\) b. \(P(t)=\ln 5 t\) c. \(P(t)=e^{t \sqrt{t}}\) d. \(\quad P(t)=e^{\sqrt{2 t}}\) e. \(P(t)=\ln (\ln t)\) f. \(
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Write a solution for each of the following derivative equations. Sketch the graph of the solution. For each, find the doubling time, \(t_{d b l},\) or half life
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Derivatives of functions are computed below. Identify the rule used in each step. In a few steps the rule is an algebraic rule of exponents and not a derivative
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(a) Compute the centered difference $$\frac{P(a+h)-P(a-h)}{2 h}$$ which is an approximation to \(P^{\prime}(a),\) for \(P(t)=t^{2}\) and compare your answer wit
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