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AveGali [126]
3 years ago
5

Using chain rule, what is the derivative of arcsin(sin(x))

Mathematics
1 answer:
oee [108]3 years ago
7 0
Chain rule:
if
y=y(u) and u=u(x)
The dy/dx=(dv/du)(du/dx)

In our case
y=arcsin(u)
u=sin(x)

dy/du=1/√(1-u²) = 1/√(1-sin²x)
du/dx=cos x

dy/dx=cos x /√(1-sin²x)

Answer: dy/dx=cos x /√(1-sin²x)
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anastassius [24]

Answer:

1) tanФ = 0.8392 ⇒ 2nd answer

2) cosФ = 20/29 ⇒ 3rd answer

3) cosФ = -√2/2 , cotФ = -1 ⇒ first answer

Step-by-step explanation:

* At first check the quadrant of the angle

∵ 180° < Ф < 270°

∴ Ф lies on the third quadrant

* Remember in the 3rd quadrant tanФ and cotФ only positive

∵ sinФ ≅ -0.7660

∵ tan²Ф = sec²Ф - 1

∵ secФ = 1/sinФ

∵ secФ = 1/-0.7660

∴ sec²Ф = (1/-0.7660)²

* Substitute this value in the equation

∴ tan²Ф = (1/-0.7660)² - 1 = 0.70428594 ⇒ take √ for both sides

∴ tanФ ≅ ± 0.8392

∵ Ф lies on the 3rd quadrant

∴ tanФ = 0.8392

* At first check the quadrant of the angle

∵ 0° < Ф < 90°

∴ Ф lies on the first quadrant

* Remember in the 1st quadrant all are positive

∵ sinФ = 21/29

∵ sin²Ф + cos²Ф = 1

∴ (21/29)² + cos²Ф = 1 ⇒ subtract (21/29)² from both sides

∴ cos²Ф = 1 - (21/29)² = 400/841 ⇒ take √ in both sides

∴ cosФ = ± 20/29

* Because Ф lies in the 1st quadrant

∴ cosФ = 20/29

* Remember that when the point is at the terminal side of angle Ф

∴ Its x-coordinate is cosФ

∴ Its y-coordinate is sinФ

* The point is (-√2/2 , √2/2)

∵ x-coordinate is negative ad y-coordinate is positive

∴ the point is on the 2nd quadrant

∴ 90° < Ф < 180°

∴ The values of sinФ and cscФ only positive

* From previous we know that:

∴ cosФ = -√2/2

∴ sinФ = √2/2

∵ cotФ = cosФ/sinФ

∴ cotФ = -√2/2 ÷ √2/2 = -√2/2 × 2/√2 = -√2/√2 = -1

* cosФ = -√2/2 , cotФ = -1

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