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Let 4x5 + y4 + 5y = 10. Compute ㅛ. dx dy dx =

Let 4x5 + y4 + 5y = 10. Compute ㅛ. dx dy dx =

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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### Implicit Differentiation Example Consider the equation \( 4x^5 + y^4 + 5y = 10 \). We are tasked with finding the derivative of \( y \) with respect to \( x \), denoted as \(\frac{dy}{dx}\). Using implicit differentiation, differentiate each term of the equation with respect to \( x \): \[ \frac{d}{dx}(4x^5) + \frac{d}{dx}(y^4) + \frac{d}{dx}(5y) = \frac{d}{dx}(10) \] First, let's differentiate \( 4x^5 \): \[ \frac{d}{dx}(4x^5) = 20x^4 \] Next, we differentiate \( y^4 \) using the chain rule: \[ \frac{d}{dx}(y^4) = 4y^3 \cdot \frac{dy}{dx} \] Similarly, differentiate \( 5y \): \[ \frac{d}{dx}(5y) = 5 \cdot \frac{dy}{dx} \] The derivative of the constant 10 is: \[ \frac{d}{dx}(10) = 0 \] Now, substituting these derivatives back into the equation, we obtain: \[ 20x^4 + 4y^3 \frac{dy}{dx} + 5 \frac{dy}{dx} = 0 \] Combine the terms involving \(\frac{dy}{dx}\): \[ 20x^4 + (4y^3 + 5)\frac{dy}{dx} = 0 \] Solve for \(\frac{dy}{dx}\): \[ (4y^3 + 5)\frac{dy}{dx} = -20x^4 \] \[ \frac{dy}{dx} = \frac{-20x^4}{4y^3 + 5} \] Finally, we obtain: \[ \frac{dy}{dx} = \boxed{\frac{-20x^4}{4y^3 + 5}} \] This represents the derivative of \( y \) with respect to \( x \) for the given equation.
Transcribed Image Text:### Implicit Differentiation Example Consider the equation \( 4x^5 + y^4 + 5y = 10 \). We are tasked with finding the derivative of \( y \) with respect to \( x \), denoted as \(\frac{dy}{dx}\). Using implicit differentiation, differentiate each term of the equation with respect to \( x \): \[ \frac{d}{dx}(4x^5) + \frac{d}{dx}(y^4) + \frac{d}{dx}(5y) = \frac{d}{dx}(10) \] First, let's differentiate \( 4x^5 \): \[ \frac{d}{dx}(4x^5) = 20x^4 \] Next, we differentiate \( y^4 \) using the chain rule: \[ \frac{d}{dx}(y^4) = 4y^3 \cdot \frac{dy}{dx} \] Similarly, differentiate \( 5y \): \[ \frac{d}{dx}(5y) = 5 \cdot \frac{dy}{dx} \] The derivative of the constant 10 is: \[ \frac{d}{dx}(10) = 0 \] Now, substituting these derivatives back into the equation, we obtain: \[ 20x^4 + 4y^3 \frac{dy}{dx} + 5 \frac{dy}{dx} = 0 \] Combine the terms involving \(\frac{dy}{dx}\): \[ 20x^4 + (4y^3 + 5)\frac{dy}{dx} = 0 \] Solve for \(\frac{dy}{dx}\): \[ (4y^3 + 5)\frac{dy}{dx} = -20x^4 \] \[ \frac{dy}{dx} = \frac{-20x^4}{4y^3 + 5} \] Finally, we obtain: \[ \frac{dy}{dx} = \boxed{\frac{-20x^4}{4y^3 + 5}} \] This represents the derivative of \( y \) with respect to \( x \) for the given equation.
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