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Question 1 of 5
Find the derivative
y=xx2+1y=xx2+1
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First, find the derivative of u and v
Derivative of v:
v |
= |
x2+1 |
v’ |
= |
2x |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdu−udvv2 |
|
y’ |
= |
(x2+1⋅1)−(x⋅2x)(x2+1)2 |
Substitute known values |
|
|
= |
x2+1-2x2(x2+1)2 |
Evaluate |
|
|
= |
1-x2(x2+1)2 |
Combine like terms |
|
|
= |
(1-x)(1+x)(x2+1)2 |
Factorize |
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Question 2 of 5
Find the derivative
y=x22x+1
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First, find the derivative of u and v
Derivative of u:
u |
= |
x2 |
u’ |
= |
2x |
Power Rule |
Derivative of v:
v |
= |
2x+1 |
v’ |
= |
2 |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdu−udvv2 |
|
y′ |
= |
(2x+1⋅2x)−(x2⋅2)(2x+1)2 |
Substitute known values |
|
|
= |
4x2+2x-2x2(2x+1)2 |
Evaluate |
|
|
= |
2x2+2x(2x+1)2 |
Combine like terms |
|
|
= |
2x(x+1)(2x+1)2 |
Factorize |
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Question 3 of 5
Find the derivative
y=x2-4x-1
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First, find the derivative of u and v
Derivative of u:
u |
= |
x2-4 |
u’ |
= |
2x |
Power Rule |
Derivative of v:
v |
= |
x-1 |
v’ |
= |
1 |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdu−udvv2 |
|
y’ |
= |
(x−1⋅2x)−(x2−4⋅1)(x−1)2 |
Substitute known values |
|
|
= |
2x(x-1)-x2+4(x-1)2 |
Evaluate |
|
|
= |
2x2-2x+x2+4(x-1)2 |
Distribute |
|
|
= |
x2-2x+4(x-1)2 |
Combine like terms |
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Question 4 of 5
Find the derivative
y=4x3-5x+3x3-1
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First, find the derivative of u and v
Derivative of u:
u |
= |
4x3-5x+3 |
u’ |
= |
12x2-5 |
Power Rule |
Derivative of v:
v |
= |
x3-1 |
v’ |
= |
3x2 |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdu−udvv2 |
|
y’ |
= |
(x3−1⋅12x2−5)−(4x3−5x+3⋅3x2)(x3−1)2 |
Substitute known values |
|
|
= |
12x5-5x3-12x2+5-12x5+15x3-9x2(x3-1)2 |
Evaluate |
|
|
= |
10x3-21x2+5(x3-1)2 |
Combine like terms |
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Question 5 of 5
Find the derivative
y=√2x+13x-2
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First, convert the surd into an exponent.
Next, find the derivative of u and v
Derivative of u:
u |
= |
(2x+1)12 |
|
u’ |
= |
(2x+1)-12 |
Chain Rule |
Derivative of v:
v |
= |
3x-2 |
v’ |
= |
3 |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdu−udvv2 |
|
y’ |
= |
(3x−2⋅(2x+1)−12)−((2x+1)12⋅3)(3x−2)2 |
Substitute known values |
|
|
= |
((3x-2)(2x+1)-12)-3((2x+1)12)(3x-2)2 |
Further simplify the numerator
|
|
((3x-2)(2x+1)-12)-3((2x+1)12) |
|
|
= |
3x-2√2x+1-3√2x+1 |
Convert the exponents into surds |
|
|
= |
(3x-2)-3(2x+1)√2x+1 |
Find the common denominator |
|
|
= |
3x-2-6x-3√2x+1 |
Distribute |
|
|
= |
-3x-5√2x+1 |
Combine like terms |
Finally, combine the simplified numerator to the denominator.
-3x-5√2x+1×1(3x-2)2 |
= |
-3x-5√2x+1(3x-2)2 |