Step 1
1 of 2
We are familiar with the given function:
g(α) = 5−α/2 sin 2α
Now we will find the derivative of the function.
We have:
g(α) = 5−α/2 sin 2α
d
d
g(α) =
(5−α/2 sin 2α)
dα
dα
d
α
α d
g ′ (α) =
(5− 2 ) sin (2α) + 5− 2
(sin (2α))
dα
dα
α
1
α d
= (5− 2 ) ln(5) (− ) ⋅ sin (2α) + 5− 2
(sin (2α))
2
dα
1
α
α
d
= − (5− 2 ) ln(5) sin (2α) + 5− 2 cos (2α)
(2α)
2
dα
α
α
1
= − (5− 2 ) ln(5) sin (2α) + 5− 2 cos(2α) ⋅ 2
2
α
1
α
= 2 ⋅ 5− 2 cos(2α) − (5− 2 ) ln(5) sin (2α)
2
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(Take the derivative of each side with r
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(Apply the Product Rule (f ⋅ g)′ = f ′
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(Apply the Logarithmic Properties
d
(ax ) =
dx
du
(Apply the Chain Rule
=
dx
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Therefore,
α
g ′ (α) = 2 ⋅ 5− 2 cos(2α) −
​
1 −α
(5 2 ) ln(5) sin (2α))
2
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