Find an equation of the tangent line to the graph of the equation at the given point. x² + x arctan y = y - 1, (- Step 1 Use implicit differentiation to find y'. 2x + (-1, 1) 4 x² + x arctan y = y - 1, + X 1 + = y' X (1 - ₁ + √²)² = 1 y' =
Find an equation of the tangent line to the graph of the equation at the given point. x² + x arctan y = y - 1, (- Step 1 Use implicit differentiation to find y'. 2x + (-1, 1) 4 x² + x arctan y = y - 1, + X 1 + = y' X (1 - ₁ + √²)² = 1 y' =
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
![Find an equation of the tangent line to the graph of the equation at the given point.
x² + x arctan y = y - 1, (-
(-1, ¹)
1
4
Step 1
Use implicit differentiation to find y'.
2x +
x² + x arctan y = y - 1,
+
X
1 + y²
(1-1
X
+y²,
y' = y'
y'](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9b6a03f-2298-4441-ab19-fb50f28ec42f%2F219953cc-7697-4c58-8144-7531370c1b4e%2Fghbvkj_processed.png&w=3840&q=75)
Transcribed Image Text:Find an equation of the tangent line to the graph of the equation at the given point.
x² + x arctan y = y - 1, (-
(-1, ¹)
1
4
Step 1
Use implicit differentiation to find y'.
2x +
x² + x arctan y = y - 1,
+
X
1 + y²
(1-1
X
+y²,
y' = y'
y'
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Follow-up Questions
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Follow-up Question
![Find an equation of the tangent line to the graph of the equation at the given point.
(-1, 1)
4
x² + x arctan y = y - 1,
Step 1
Use implicit differentiation to find y'.
Step 3
2x +
y' =
m =
arctan y
m =
b =
π
atan (y)
(1 + y²)(-2x - arctan y)
x - y² - 1
-2π
π + 8
+arctan(1)
π
Using the slope and the given point
x² + x arctan y = y - 1,
Step 2
From the derivative, find the slope of the tangent line at the given point (- , 1).
π
4
1+ (1)²
1
X
+
X
1 + y2V'
X
1+ y²,
= y'
-2πT
8+T
y' =
2x + arctan y
2x + arctan y
X
1
1+y²
2x + atan (y)
2x + atan (y)
1-(1)
(-1). find the y-intercept of the line tangent to the given point.](https://content.bartleby.com/qna-images/question/f9b6a03f-2298-4441-ab19-fb50f28ec42f/a5a0bbc1-8753-4bcb-bdac-4c04731ba646/y2pataj_thumbnail.png)
Transcribed Image Text:Find an equation of the tangent line to the graph of the equation at the given point.
(-1, 1)
4
x² + x arctan y = y - 1,
Step 1
Use implicit differentiation to find y'.
Step 3
2x +
y' =
m =
arctan y
m =
b =
π
atan (y)
(1 + y²)(-2x - arctan y)
x - y² - 1
-2π
π + 8
+arctan(1)
π
Using the slope and the given point
x² + x arctan y = y - 1,
Step 2
From the derivative, find the slope of the tangent line at the given point (- , 1).
π
4
1+ (1)²
1
X
+
X
1 + y2V'
X
1+ y²,
= y'
-2πT
8+T
y' =
2x + arctan y
2x + arctan y
X
1
1+y²
2x + atan (y)
2x + atan (y)
1-(1)
(-1). find the y-intercept of the line tangent to the given point.
Solution
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