Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. arctan(x + y) = y² + I, (1, 0) 4
Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. arctan(x + y) = y² + I, (1, 0) 4
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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![Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point.
π
4
(1, 0)
Step 1
To find the equation of the tangent line to the graph of arctan(x + y) = y² + differentiate arctan(x + y) = y² + with respect to x. Therefore,
-I
4
π
Step 2
arctan(x + y) = y² +
Submit
d
1
1 + x + y
1 + 1
(arctan(x + y))
1
[1 + y']
1
To find the value of y' at the given point (1, 0), substitute x = 1 and y = 0 in
+ 0)² [1
=
Skin (you cannot come back)
=
2yy'.
[1 + y'] = 20
-[1 + y' ] =
=
+
⇒ y' =
11)
y'
1
1 + (x + y)²
-[1 + y'] = 2yy' and solve for y'. Therefore,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9b6a03f-2298-4441-ab19-fb50f28ec42f%2F288d6df0-5ad6-4e42-8c12-9f7dc1ea18b0%2Fkipju1p_processed.png&w=3840&q=75)
Transcribed Image Text:Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point.
π
4
(1, 0)
Step 1
To find the equation of the tangent line to the graph of arctan(x + y) = y² + differentiate arctan(x + y) = y² + with respect to x. Therefore,
-I
4
π
Step 2
arctan(x + y) = y² +
Submit
d
1
1 + x + y
1 + 1
(arctan(x + y))
1
[1 + y']
1
To find the value of y' at the given point (1, 0), substitute x = 1 and y = 0 in
+ 0)² [1
=
Skin (you cannot come back)
=
2yy'.
[1 + y'] = 20
-[1 + y' ] =
=
+
⇒ y' =
11)
y'
1
1 + (x + y)²
-[1 + y'] = 2yy' and solve for y'. Therefore,
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