Respuesta :

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