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Question 1 of 5
Find the derivative
y=x4e−x
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First, find the derivative of u and v
Derivative of u:
u |
= |
x4 |
u’ |
= |
4x3 |
Derivative of v:
v |
= |
e−x |
v’ |
= |
−e−x |
Substitute the components into the product rule
dydx |
= |
vdudx+udvdx |
|
y′ |
= |
(e−x⋅4x3)+(x4⋅−e−x) |
Substitute known values |
y’ |
= |
4x3e−x−x4e−x |
Evaluate |
y’=4x3e−x−x4e−x
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Question 2 of 5
Find the derivative
y=(x2−x)e2x−1
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First, find the derivative of u and v
Derivative of u:
u |
= |
x2−x |
u’ |
= |
2x−1 |
Derivative of v:
v |
= |
e2x−1 |
v’ |
= |
2e2x−1 |
Substitute the components into the product rule
dydx |
= |
vdudx+udvdx |
|
y′ |
= |
(e2x−1⋅(2x−1))+((x2−x)⋅2e2x−1) |
Substitute known values |
|
= |
2xe2x−1−e2x−1+2x2e2x−1−2xe2x−1 |
Evaluate |
|
= |
−e2x−1+2x2e2x−1 |
2xe2x−1−2xe2x−1 cancels out |
y’=−e2x−1+2x2e2x−1
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Question 3 of 5
Find the derivative
y=3−x2e2
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First, find the derivative of u and v
Derivative of u:
u |
= |
3−x2 |
u’ |
= |
−2x |
Substitute the components into the product rule
dydx |
= |
vdudx−udvdxv2 |
|
y′ |
= |
(ex⋅−2x)−((3−x2)⋅ex)(ex)2 |
Substitute known values |
|
|
= |
−2xex−[(3−x2)ex]e2x |
Evaluate |
y’=−2xex−[(3−x2)ex]e2x
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Question 4 of 5
Find the derivative
y=ex−1ex+1
Incorrect
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First, find the derivative of u and v
Derivative of u:
u |
= |
ex−1 |
u’ |
= |
ex |
Derivative of v:
v |
= |
ex+1 |
v’ |
= |
ex |
Substitute the components into the product rule
dydx |
= |
vdudx−udvdxv2 |
|
y′ |
= |
((ex+1)⋅ex)−((ex−1)⋅ex)(ex+1)2 |
Substitute known values |
|
|
= |
e2x+ex−e2x+ex(ex+1)2 |
Evaluate |
|
|
= |
2ex(ex+1)2 |
Simplify |
y’=2ex(ex+1)2
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Question 5 of 5
Find the derivative
y=√e2x+7
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First, find the derivative of u and the dervative of y with respect to u.
Derivative of u:
u |
= |
e2x+7 |
|
dudx |
= |
2e2x |
Derivative of y with respect to u:
y |
= |
u12 |
|
dydu |
= |
12(e2x+7)−12 |
Substitute the components into the chain rule
dydx |
= |
dydu⋅dudx |
|
y′ |
= |
12(e2x+7)−12⋅2e2x |
Substitute known values |
|
|
= |
e2x(e2x+7)−12 |
12×2=1 |
|
|
= |
e2x(e2x+7)12 |
Reciprocate (e2x+7)−12 |
|
|
= |
e2x√e2x+7 |
Change the exponent to a surd |