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Sagot :
Answer:csc
2
(
x
)
=
1
sin
2
(
x
)
Step-by-step explanation:
d
d
x
[
csc
2
(
x
)
]
=
−
2
cot
x
csc
2
x
Explanation:
csc
2
(
x
)
=
1
sin
2
(
x
)
d
d
x
[
csc
2
(
x
)
]
=
d
d
x
[
1
sin
2
(
x
)
]
d
d
x
[
1
sin
2
(
x
)
]
=
d
d
x
[
[
sin
(
x
)
]
−
2
]
let
u
=
sin
x
d
d
x
[
[
sin
(
x
)
]
−
2
]
=
d
d
u
[
u
−
2
]
d
d
x
[
sin
x
]
d
d
u
[
u
−
2
]
=
−
2
u
−
3
d
d
x
[
sin
x
]
=
cos
x
d
d
x
[
[
sin
(
x
)
]
−
2
]
=
−
2
u
−
3
cos
x
=
−
2
cos
x
sin
3
x
cos
x
sin
x
=
cot
x
⇒
−
2
cos
x
sin
3
x
=
−
2
cot
x
sin
2
x
1
sin
2
x
=
csc
2
x
⇒
−
2
cot
x
csc
2
x
d
d
x
[
csc
2
(
x
)
]
=
−
2
cot
x
csc
2
x
The derivative of cosec x² is [tex]-2x.cot \ x^{2} .cosec \ x^{2}[/tex]
Differentiation
From the question, we are to differentiate cosec x²
Let y = cosec x², then we will determine the value of [tex]\frac{dy}{dx}[/tex]
[tex]y = cosec \ x^{2}[/tex]
Let [tex]u = x^{2}[/tex]
Then,
[tex]\frac{du}{dx}=2x[/tex]
Also, [tex]y = cosec \ x^{2}[/tex] becomes
[tex]y = cosec \ u[/tex]
Then,
[tex]\frac{dy}{du} = - cot \ u . cosec \ u[/tex]
From Chain rule
[tex]\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}[/tex]
∴ [tex]\frac{dy}{dx} = -cot \ u.cosec \ u \times 2x[/tex]
Recall that [tex]u = x^{2}[/tex]
Then,
[tex]\frac{dy}{dx} = -cot \ x^{2} .cosec \ x^{2} \times 2x[/tex]
[tex]\frac{dy}{dx} = -2x.cot \ x^{2} .cosec \ x^{2}[/tex]
Hence, the derivative of cosec x² is [tex]-2x.cot \ x^{2} .cosec \ x^{2}[/tex]
Learn more on Differentiation here: https://brainly.com/question/25081524
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