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Find dy/dx by implicit differentiation. x4 + yA 5

Find dy/dx by implicit differentiation. x4 + yA 5

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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Find dy/dx by implicit differentiation. x4 +yA = 5 Part 1 of 3 dy - = y' by implicit differentiation, we consider y to be a function of x. This means that When finding xp whenever a derivative is calculated for an expression that includes the variable y, the Chain Rule requires that [v2] = (2y)(y'). we multiply the derivative by y'. For example, xp Similarly, 4y3 D(v). %3D xp Part 2 of 3 To find = y' for x4 + y = 5, we differentiate both sides of the equation with respect to x. dy dx On the right-hand side, we have xp d On the left-hand side, we have [x + y*]. We already know that [y*] = 4y³y'. For the term x, we dx dx have d %3D dx
Transcribed Image Text:Find dy/dx by implicit differentiation. x4 +yA = 5 Part 1 of 3 dy - = y' by implicit differentiation, we consider y to be a function of x. This means that When finding xp whenever a derivative is calculated for an expression that includes the variable y, the Chain Rule requires that [v2] = (2y)(y'). we multiply the derivative by y'. For example, xp Similarly, 4y3 D(v). %3D xp Part 2 of 3 To find = y' for x4 + y = 5, we differentiate both sides of the equation with respect to x. dy dx On the right-hand side, we have xp d On the left-hand side, we have [x + y*]. We already know that [y*] = 4y³y'. For the term x, we dx dx have d %3D dx
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