Explanation Given: The equation is: 5 − y 2 = 6 − 2 x 4 . And the point is ( 1 , 1 ) . Formula used: The slope-point form of the equation of a line with the slope m is: y − y 1 = m ( x − x 1 ) And, d d x ( u ⋅ v ) = u d v d x + v d u d x d d x x n = n x n − 1 d d x e x = e x Calculation: Consider the provided equation of the curve, 5 − y 2 = 6 − 2 x 4 Differentiate both sides of the above equation with respect to x , d d x ( 5 ) − d d x ( y 2 ) = d d x ( 6 ) − d d x ( 2 x 4 ) 0 − 2 y d y d x = <

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

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ISBN: 9781323751671

Author: Lial

Publisher: PEARSON C

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Chapter 12, Problem 46RE

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To calculate: The equation of tangent line to the curve, 5−y2=6−2x4 at the point (1,1).

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Chapter 12, Problem 45RE

Chapter 12, Problem 47RE

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Chapter 12 Solutions

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

Ch. 12.1 - Checkpoint 1 For what values of x is the function...Ch. 12.1 - Checkpoint 2 Find all intervals on which is...Ch. 12.1 - Checkpoint 3 Identity the x-values of all points...Ch. 12.1 - Checkpoint 4 Find the critical numbers for each of...Ch. 12.1 - Prob. 5CP Ch. 12.1 - Prob. 6CP Ch. 12.1 - Checkpoint 7 Find the locations of the local...Ch. 12.1 - Prob. 8CP Ch. 12.1 - Checkpoint 9 If a sales function is given by...Ch. 12.1 - Prob. 1E

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The equation of tangent line to the curve, 5 − y 2 = 6 − 2 x 4 at the point ( 1 , 1 ) .

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To calculate: The equation of tangent line to the curve, 5−y2=6−2x4 at the point (1,1).

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Chapter 12 Solutions