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Exam (elaborations)

Math 225 Week 4 Test Summer 2025

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Question 1 : 5 ptsSkip to question text. Given the equation2x+4y=8,find dydx.Given the equation2x+4y=8,find dydx. dydx=−32dydx=−32 dydx=32dydx=32 dydx=−12dydx=−12 dydx=12dydx=12 2x + 4y = 8 5 / 5 Question 2 : 5 ptsSkip to question text. Suppose a curve is defined by the equation(6−x)y2=x3. What is the equation of theline tangent to the curv e at (3, 3)? Suppose a curve is defined by the equation(6−x)y2=x3. What is the equation of theline tangent to the curve at (3, 3)? y=2x+3y=2x+3 y=2x−3y=2x−3 y=12x+32y=12x+32 y=12x−32y=12x−32 (6−x)y2=x36y2−xy2=x3ddx6y2−ddxxy2=ddxx312ydydx−2xydydx−y2=3x212ydydx−2xydydx=3x2+y2dy

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Week 4 Test - Grade: 90%
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Questions Limits Points Due Date


20 Questions 180 Minutes 100 pts possible No due date.




Attempt 1 90% (90 of 100) Completed on 03/30/25 at 09:18PM
Score for this quiz: 90% ( 90 /100)
Submitted Mar 30 at 9:18pm
This attempt took 36 minutes.

Question 1 : 5 ptsSkip to question text.
Given the equation2x+4y=8,find dydx.Given the equation2x+4y=8,find dydx.
dydx=−32dydx=−32
dydx=32dydx=32
dydx=−12dydx=−12
dydx=12dydx=12

2x +
4y = 8

, d d
(2x + (8
d 4y) = d ) 5 /5
x x
d d d
Question 2 : 5 ptsSkip to question text.
2x 4y (8 Suppose a curve is defined by the equation(6−x)y2=x3. What is the equation of theline tangent to the curv
d + d = d )
e at (3, 3)?
x x x Suppose a curve is defined by the equation(6−x)y2=x3. What is the equation of theline tangent to the curve at (3, 3)?
d d d y=2x+3y=2x+3
x+ y (8 y=2x−3y=2x−3
2
d 4 d = d ) y=12x+32y=12x+32
x x x y=12x−32y=12x−32
d (6−x)y2=x36y2−xy2=x3ddx6y2−ddxxy2=ddxx312ydydx−2xydydx−y2=3x212ydydx−2xydydx=3x2+y2dy
y dx(12y−2xy)=3x2+y2dydx=3x2+y212y−2xym=3(3)2+(3)212(3)−2(3)
2+ =
4 0 (3)=2y−y0=m(x−x0)y=2x−3(6−x)y2=x36y2−xy2=x3ddx6y2−ddxxy2=ddxx312ydydx−2xydydx−y2=3x21
d
2ydydx−2xydydx=3x2+y2dydx(12y−2xy)=3x2+y2dydx=3x2+y212y−2xym=3(3)2+(3)212(3)−2(3)
x
(3)=2y−y0=m(x−x0)y=2x−3
d
5 /5
y
=
4
−2
d Question 3 : 5 ptsUse the linear approximation formula to approximate e 2.1.
x Use the linear approximation formula to approximate e 2.1.
d e2.1≈8.1661699e2.1≈8.1661699
− −
y e2.1≈11e210e2.1≈11e210
2 1
= = . e2.1≈e2e2.1≈e2
d e2.1≈9e210e2.1≈9e210
4 2
x According to the linear approximation formula,f(x0+Δx)≈f′(x0)(Δx)
+f(x0).To use the formula, you will need to know thefirst derivative.f′(x)=exTherefore:f(x0+Δx)≈f′(x0)(Δx)
+f(x0)f(2+.1)≈f′(2)(.1)+f(2)f(2.1)≈e2(.1)+e2f(2.1)≈e2+10e210f(2.1)=e2.1≈11e210According to the linear approximation f
ormula,f(x0+Δx)≈f′(x0)(Δx)+f(x0).To use the formula, you will need to know thefirst derivative.f′
(x)=exTherefore:f(x0+Δx)≈f′(x0)(Δx)+f(x0)f(2+.1)≈f′(2)(.1)+f(2)f(2.1)≈e2(.1)+e2f(2.1)≈e2+10e210f(2.1)=e2.1≈11e210
5 /5


Question 4 : 5 ptsGiven cos (x + y) = sin x sin y, find dy / dx by implicit differentiation.
Given cos (x + y) = sin x sin y, find dy / dx by implicit differentiation.
sin(x+y)+cosxsinysin(x+y)+sinxcosysin(x+y)+cosxsinysin(x+y)+sinxcosy
−cos(x+y)+cosxsinycos(x+y)+sinxcosy−cos(x+y)+cosxsinycos(x+y)+sinxcosy
−sin(x+y)+cosxsinysin(x+y)+sinxcosy−sin(x+y)+cosxsinysin(x+y)+sinxcosy
sin(x+y)−cosxsinysin(x+y)−sinxcosysin(x+y)−cosxsinysin(x+y)−sinxcosy

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