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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Question
Can you explain what is happening to the red circled part of this problem?
thanks
![Chapter 3.5, Problem 18E
To determine
To calculate: The dy of function (sin tæ — cos
Expert Solution & Answer
Answer to Problem 18E
dy
The of function (sin x + cos y)² = 2 by implicit differentiation is d
dx
Explanation of Solution
Given:
The function is:
(sin πx - cos πy)² = 2
Formula used:
The derivative of function y = sin ax with respect to a is;
= a cos x
dr
The derivative of function y = a cos x with respect to a is;
dy
dr
= -a sin æ
Calculation:
Use the implicit differentiation by differentiating with respect to x as;
d
(sin л сos TY)
sny) = 4 (2)
dx
Now, the derivative of function is:
2 (sin πx + cos Ty)
Simplified
= -2Y no?
2 (sin Tacos Ty) COS TX -
The
dx
TT COS TX
sin Ty
COS T
sin Ty
[(sin TX + COS TY)] = = 0
(2)]=
dr
π COS TX - Tπ sin
Further calculation shows that:
dy
dx
₁ πy ( 1² ) =
os πy)² = 2 by implicit differentiation.
π sin Ty
πως πX = π sin my
dr
= 0
= 0
of function (sin mx + cos y)² = 2 by implicit differentiation is
dy
dr
=
COS TX
sin Ty
COS TX
sin Ty
SAVE](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd77e6217-e8e9-4b82-a244-bb8b4b6e855a%2Fd4f4cc38-c6fc-4103-a3e9-c6aa13a2e0e0%2Fv86bm5c_processed.png&w=3840&q=75)
Transcribed Image Text:Chapter 3.5, Problem 18E
To determine
To calculate: The dy of function (sin tæ — cos
Expert Solution & Answer
Answer to Problem 18E
dy
The of function (sin x + cos y)² = 2 by implicit differentiation is d
dx
Explanation of Solution
Given:
The function is:
(sin πx - cos πy)² = 2
Formula used:
The derivative of function y = sin ax with respect to a is;
= a cos x
dr
The derivative of function y = a cos x with respect to a is;
dy
dr
= -a sin æ
Calculation:
Use the implicit differentiation by differentiating with respect to x as;
d
(sin л сos TY)
sny) = 4 (2)
dx
Now, the derivative of function is:
2 (sin πx + cos Ty)
Simplified
= -2Y no?
2 (sin Tacos Ty) COS TX -
The
dx
TT COS TX
sin Ty
COS T
sin Ty
[(sin TX + COS TY)] = = 0
(2)]=
dr
π COS TX - Tπ sin
Further calculation shows that:
dy
dx
₁ πy ( 1² ) =
os πy)² = 2 by implicit differentiation.
π sin Ty
πως πX = π sin my
dr
= 0
= 0
of function (sin mx + cos y)² = 2 by implicit differentiation is
dy
dr
=
COS TX
sin Ty
COS TX
sin Ty
SAVE
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