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For the equation given below, evaluate y at the point (2, −1). y'at (2, -1) = e³ + 10 − e¯¹ = 2x² + 2y².

For the equation given below, evaluate y at the point (2, −1). y'at (2, -1) = e³ + 10 − e¯¹ = 2x² + 2y².

Calculus: Early Transcendentals
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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For the equation given below, evaluate \( y' \) at the point \( (2, -1) \). \[ e^y + 10 - e^{-1} = 2x^2 + 2y^2 \] \[ y' \text{ at } (2, -1) = \, \]
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:** \(\frac{dy}{dx} = -\frac{4x^7}{y}\)
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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