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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
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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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'
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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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.
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.
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