Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Concept explainers
Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question
![**Problem 8:**
Find \(\frac{dy}{dt}\) if \( y = 3e^t - 2^{10} - t^{\frac{\pi}{2}} + e^{\sqrt{5}} + \frac{1}{\sqrt{t}} \). Leave the answer without negative exponents.
**Solution Overview:**
To find \(\frac{dy}{dt}\), we need to differentiate each term of the function \(y\) with respect to \(t\):
1. \( 3e^t \): Differentiate using the chain rule; derivative is \(3e^t\).
2. \(-2^{10}\): This is a constant, so its derivative is 0.
3. \(-t^{\frac{\pi}{2}}\): Use the power rule; derivative is \(-\frac{\pi}{2}t^{\frac{\pi}{2} - 1}\).
4. \(e^{\sqrt{5}}\): This is a constant (since \(\sqrt{5}\) is not a function of \(t\)), so its derivative is 0.
5. \(\frac{1}{\sqrt{t}}\): Rewrite as \(t^{-\frac{1}{2}}\) and use the power rule; derivative is \(-\frac{1}{2}t^{-\frac{3}{2}}\).
Finally, combine the differentiated terms:
\[
\frac{dy}{dt} = 3e^t - \frac{\pi}{2}t^{\frac{\pi}{2} - 1} - \frac{1}{2t^{\frac{3}{2}}}
\]
Convert \(\frac{1}{2t^{\frac{3}{2}}}\) to non-negative exponents:
\[
\frac{dy}{dt} = 3e^t - \frac{\pi}{2}t^{\frac{\pi}{2} - 1} - \frac{1}{2t\sqrt{t}}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26e4b470-d78e-41a3-856d-1504f09d7f45%2Fff1c424c-c38c-4ed4-a638-ee6f008fb9bb%2Fi8bu9w6_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 8:**
Find \(\frac{dy}{dt}\) if \( y = 3e^t - 2^{10} - t^{\frac{\pi}{2}} + e^{\sqrt{5}} + \frac{1}{\sqrt{t}} \). Leave the answer without negative exponents.
**Solution Overview:**
To find \(\frac{dy}{dt}\), we need to differentiate each term of the function \(y\) with respect to \(t\):
1. \( 3e^t \): Differentiate using the chain rule; derivative is \(3e^t\).
2. \(-2^{10}\): This is a constant, so its derivative is 0.
3. \(-t^{\frac{\pi}{2}}\): Use the power rule; derivative is \(-\frac{\pi}{2}t^{\frac{\pi}{2} - 1}\).
4. \(e^{\sqrt{5}}\): This is a constant (since \(\sqrt{5}\) is not a function of \(t\)), so its derivative is 0.
5. \(\frac{1}{\sqrt{t}}\): Rewrite as \(t^{-\frac{1}{2}}\) and use the power rule; derivative is \(-\frac{1}{2}t^{-\frac{3}{2}}\).
Finally, combine the differentiated terms:
\[
\frac{dy}{dt} = 3e^t - \frac{\pi}{2}t^{\frac{\pi}{2} - 1} - \frac{1}{2t^{\frac{3}{2}}}
\]
Convert \(\frac{1}{2t^{\frac{3}{2}}}\) to non-negative exponents:
\[
\frac{dy}{dt} = 3e^t - \frac{\pi}{2}t^{\frac{\pi}{2} - 1} - \frac{1}{2t\sqrt{t}}
\]
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