dy -2x² + 5y5 + ey = 0. Compute in terms of x and y. dx Let y(x) be defined implicitly by G(x, y(x)) = (Use symbolic notation and fractions where needed.) dy dx ||
dy -2x² + 5y5 + ey = 0. Compute in terms of x and y. dx Let y(x) be defined implicitly by G(x, y(x)) = (Use symbolic notation and fractions where needed.) dy dx ||
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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![### Implicit Differentiation
**Problem Statement:**
Let \( y(x) \) be defined implicitly by \( G(x, y(x)) = -2x^7 + 5y^5 + e^y = 0 \). Compute \(\dfrac{dy}{dx}\) in terms of \(x\) and \(y\).
*Use symbolic notation and fractions where needed.*
\[ \frac{dy}{dx} = \underline{\hspace{4cm}} \]
**Detailed Explanation:**
To find \(\frac{dy}{dx}\), we will differentiate the given function implicitly with respect to \(x\).
Given:
\[ G(x, y(x)) = -2x^7 + 5y^5 + e^y = 0 \]
We need to compute \(\frac{dy}{dx}\).
1. Differentiate both sides of the equation with respect to \(x\).
\[
\frac{d}{dx} \left( -2x^7 + 5y^5 + e^y \right) = \frac{d}{dx} (0)
\]
2. Apply the chain rule:
\[
\frac{d}{dx}(-2x^7) + \frac{d}{dx} (5y^5) + \frac{d}{dx}(e^y) = 0
\]
This becomes:
\[
-14x^6 + 25y^4 \frac{dy}{dx} + e^y \frac{dy}{dx} = 0
\]
3. Isolate \(\frac{dy}{dx}\):
\[
25y^4 \frac{dy}{dx} + e^y \frac{dy}{dx} = 14x^6
\]
\[
\left(25y^4 + e^y \right) \frac{dy}{dx} = 14x^6
\]
\[
\frac{dy}{dx} = \frac{14x^6}{25y^4 + e^y}
\]
Finally,
\[
\frac{dy}{dx} = \frac{14x^6}{25y^4 + e^y}
\]
Plug this result into the provided box to complete the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ec2020e-61d4-433f-ae87-5496300bb3a2%2Fe460db77-cb97-4250-9b47-b69ba81c5a8d%2Fi7ik7il_processed.png&w=3840&q=75)
Transcribed Image Text:### Implicit Differentiation
**Problem Statement:**
Let \( y(x) \) be defined implicitly by \( G(x, y(x)) = -2x^7 + 5y^5 + e^y = 0 \). Compute \(\dfrac{dy}{dx}\) in terms of \(x\) and \(y\).
*Use symbolic notation and fractions where needed.*
\[ \frac{dy}{dx} = \underline{\hspace{4cm}} \]
**Detailed Explanation:**
To find \(\frac{dy}{dx}\), we will differentiate the given function implicitly with respect to \(x\).
Given:
\[ G(x, y(x)) = -2x^7 + 5y^5 + e^y = 0 \]
We need to compute \(\frac{dy}{dx}\).
1. Differentiate both sides of the equation with respect to \(x\).
\[
\frac{d}{dx} \left( -2x^7 + 5y^5 + e^y \right) = \frac{d}{dx} (0)
\]
2. Apply the chain rule:
\[
\frac{d}{dx}(-2x^7) + \frac{d}{dx} (5y^5) + \frac{d}{dx}(e^y) = 0
\]
This becomes:
\[
-14x^6 + 25y^4 \frac{dy}{dx} + e^y \frac{dy}{dx} = 0
\]
3. Isolate \(\frac{dy}{dx}\):
\[
25y^4 \frac{dy}{dx} + e^y \frac{dy}{dx} = 14x^6
\]
\[
\left(25y^4 + e^y \right) \frac{dy}{dx} = 14x^6
\]
\[
\frac{dy}{dx} = \frac{14x^6}{25y^4 + e^y}
\]
Finally,
\[
\frac{dy}{dx} = \frac{14x^6}{25y^4 + e^y}
\]
Plug this result into the provided box to complete the problem.
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