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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Question
![For the equation given below, evaluate \( y' \) at the point \( (2, -1) \).
\[
e^y + 10 - e^{-1} = 2x^2 + 2y^2
\]
\[
y' \text{ at } (2, -1) = \,
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93dc20bf-0fea-4660-8028-43a1db35511d%2Feca9313b-9bf1-4b09-98f7-3278fc4651c1%2F3iwkhmh.png&w=3840&q=75)
Transcribed Image Text:For the equation given below, evaluate \( y' \) at the point \( (2, -1) \).
\[
e^y + 10 - e^{-1} = 2x^2 + 2y^2
\]
\[
y' \text{ at } (2, -1) = \,
\]
![**Problem Statement:**
Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) if \(x^8 + y^2 = \sqrt{5}\).
**Solution:**
To solve this problem, we need to differentiate the given equation with respect to \(x\).
Given equation:
\[ x^8 + y^2 = \sqrt{5} \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^8) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\sqrt{5}) \]
The derivative of \(x^8\) with respect to \(x\) is \(8x^7\).
The derivative of \(y^2\) with respect to \(x\) is \(2y \cdot \frac{dy}{dx}\) (using the chain rule).
The derivative of \(\sqrt{5}\) with respect to \(x\) is \(0\) (since it's a constant).
Thus, the differentiated equation is:
\[ 8x^7 + 2y \cdot \frac{dy}{dx} = 0 \]
Now, solve for \(\frac{dy}{dx}\):
\[ 2y \cdot \frac{dy}{dx} = -8x^7 \]
\[ \frac{dy}{dx} = -\frac{8x^7}{2y} \]
Simplify:
\[ \frac{dy}{dx} = -\frac{4x^7}{y} \]
**Final Answer:**
\(\frac{dy}{dx} = -\frac{4x^7}{y}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93dc20bf-0fea-4660-8028-43a1db35511d%2Feca9313b-9bf1-4b09-98f7-3278fc4651c1%2F4z4z7ur_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) if \(x^8 + y^2 = \sqrt{5}\).
**Solution:**
To solve this problem, we need to differentiate the given equation with respect to \(x\).
Given equation:
\[ x^8 + y^2 = \sqrt{5} \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^8) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\sqrt{5}) \]
The derivative of \(x^8\) with respect to \(x\) is \(8x^7\).
The derivative of \(y^2\) with respect to \(x\) is \(2y \cdot \frac{dy}{dx}\) (using the chain rule).
The derivative of \(\sqrt{5}\) with respect to \(x\) is \(0\) (since it's a constant).
Thus, the differentiated equation is:
\[ 8x^7 + 2y \cdot \frac{dy}{dx} = 0 \]
Now, solve for \(\frac{dy}{dx}\):
\[ 2y \cdot \frac{dy}{dx} = -8x^7 \]
\[ \frac{dy}{dx} = -\frac{8x^7}{2y} \]
Simplify:
\[ \frac{dy}{dx} = -\frac{4x^7}{y} \]
**Final Answer:**
\(\frac{dy}{dx} = -\frac{4x^7}{y}\)
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