vative of y with respect to x, the Chain Rule will be required and produce a factor that dx late. Here, we will also use the Product Rule on the left-hand side. d dx (52*) + d dx d dx (5x² + 3x²y — 7xy¹) (3x²y) - d dx = d dx (7xy¹) = 0 (-2)

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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How the +- signs change in the circled steps?

Differentiating the equation implicitly with respect to x, we will treat y as a function of x. When we take
dy
the derivative of y with respect to x, the Chain Rule will be required and produce a factor that we will
dx
then isolate. Here, we will also use the Product Rule on the left-hand side.
2 dy
dx
20x³ + 3x².
d
dx
(52*) +
d
+y (3x²)
dx
20x³ + 3x².
-
d
dx
3
20x³ 3x²
d
dx
dy
dx
d
(5x+3x²y-7xy¹)
(−2)
dx
(3x²y) (7xy¹) = 0
d
d
4
7x- (y¹) + yª (7x)
dx
dx
dy
+ y (6x) - (7x (4y³) +y² (7))
dx
dy
dx
d
dx
3x²
3
+ 6xy · 28xy³.
dy
dx
dy
dx
28xy ³ dy
dx
=
=
dy
dx
7y¹ = 0
= 0
-20x³ - 6x + 7y¹
3
(d) (3x² − 28ay³) = − 20x³ + 7y¹ − 6xy
0
=
=
-20x³ + 7y¹ - 6xy
3x² - 28xy³
Transcribed Image Text:Differentiating the equation implicitly with respect to x, we will treat y as a function of x. When we take dy the derivative of y with respect to x, the Chain Rule will be required and produce a factor that we will dx then isolate. Here, we will also use the Product Rule on the left-hand side. 2 dy dx 20x³ + 3x². d dx (52*) + d +y (3x²) dx 20x³ + 3x². - d dx 3 20x³ 3x² d dx dy dx d (5x+3x²y-7xy¹) (−2) dx (3x²y) (7xy¹) = 0 d d 4 7x- (y¹) + yª (7x) dx dx dy + y (6x) - (7x (4y³) +y² (7)) dx dy dx d dx 3x² 3 + 6xy · 28xy³. dy dx dy dx 28xy ³ dy dx = = dy dx 7y¹ = 0 = 0 -20x³ - 6x + 7y¹ 3 (d) (3x² − 28ay³) = − 20x³ + 7y¹ − 6xy 0 = = -20x³ + 7y¹ - 6xy 3x² - 28xy³
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