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Question 1 of 5
Find the derivative using the product rule
tan3xcosx
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Derivatives of Trigonometric Functions
Product Rule
dydx=vdudx+udvdx
First, find the derivative of u and v
Derivative of u:
u |
= |
tan 3x |
u’ |
= |
dudx |
= |
3 sec23x |
Product Rule |
Derivative of v:
v |
= |
cos x |
v’ |
= |
dvdx |
= |
-sin x |
Substitute the components into the product rule
dydx |
= |
vdudx+udvdx |
|
dydx |
= |
(cosx⋅3sec23x)+[tan3x⋅(−sinx)] |
Substitute known values |
y’ |
= |
3cos 3x sec2 3x-sin x tan 3x |
Evaluate |
y’=3cos 3x sec2 3x-sin x tan 3x
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Question 2 of 5
Find the derivative using the product rule
8xsin(x4)
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Derivatives of Trigonometric Functions
Product Rule
dydx=vdudx+udvdx
First, find the derivative of u and v
Derivative of u:
u |
= |
8x |
u’ |
= |
dudx |
= |
8 |
Power Rule |
Derivative of v:
v |
= |
sin x4 |
v’ |
= |
dvdx |
= |
14cos x4 |
Power Rule |
Substitute the components into the product rule
dydx |
= |
vdudx+udvdx |
|
dydx |
= |
(sinx4⋅8)+(8x⋅14cosx4) |
Substitute known values |
|
y’ |
= |
8sin x4+2xcos x4 |
Evaluate |
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Question 3 of 5
Find the derivative using the product rule
x2cos3x
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Derivatives of Trigonometric Functions
Product Rule
dydx=vdudx+udvdx
First, find the derivative of u and v
Derivative of u:
u |
= |
x2 |
u’ |
= |
dudx |
= |
2x |
Power Rule |
Derivative of v:
v |
= |
(cos x)3 |
v’ |
= |
dvdx |
= |
-3cos2 x sin x |
Chain Rule |
Substitute the components into the product rule
dydx |
= |
vdudx+udvdx |
|
dydx |
= |
(cos3x⋅2x)+(x2⋅−3cos2xsinx) |
Substitute known values |
|
y’ |
= |
2xcos3 x-3x2cos2x sin x |
Evaluate |
y’=2xcos3 x-3x2cos2x sin x
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Question 4 of 5
Find the derivative using the quotient rule
tanxx
Correct
div class=”s-correct”>
Fantastic!
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Derivatives of Trigonometric Functions
Quotient Rule
dydx=vdudx−udvdxv2
First, find the derivative of u and v
Derivative of u:
u |
= |
tan x |
u’ |
= |
dudx |
= |
sec2 x |
Derivative of v:
v |
= |
x |
v’ |
= |
dvdx |
= |
1 |
Power Rule |
Substitute the components into the quotient rule
dydx |
= |
vdudx−udvdxv2 |
|
dydx |
= |
(x⋅sec2x)−(tanx⋅1)x2 |
Substitute known values |
|
y’ |
= |
x sec2 x-tan xx2 |
Evaluate |
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Question 5 of 5
Find the derivative using the quotient rule
sin3xcos4x
Incorrect
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Derivatives of Trigonometric Functions
Quotient Rule
dydx=vdudx−udvdxv2
First, find the derivative of u and v
Derivative of u:
u |
= |
sin 3x |
u’ |
= |
dudx |
= |
3cos 3x |
Chain Rule |
Derivative of v:
v |
= |
cos 4x |
v’ |
= |
dvdx |
= |
-4sin 4x |
Chain Rule |
Substitute the components into the quotient rule
dydx |
= |
vdudx−udvdxv2 |
|
dydx |
= |
(cos4x⋅3cos3x)−[sin3x⋅(−4sin4x)]cos4x2 |
Substitute known values |
|
y’ |
= |
3cos 4x cos3x+4sin 3x sin 4x(cos4x)2 |
Evaluate |
y’=3cos 4x cos3x+4sin 3x sin 4x(cos4x)2