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# derivative of cosec x^2 by first principle?

### 4 Answers

- PhilipLv 64 weeks ago
[csc(x)]^2 = s^(-2), where [s,c] = [sin(x), cos(x)].

{d/dx}[s^(-2)] = -2s^(-3)*(ds/dx) = -2cs^(-3).

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- MyRankLv 64 weeks ago
cosec²x

d/dx(cosec²x)

= -2cosecx.cosecx.cotx

= -2cosec²xcotx

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- 4 weeks ago
If that's csc(x)^2, then it's just a bit of algebra to find the derivative. If it's csc(x^2), then that's a horse of a different color.

f(x) = csc(x)^2

f(x + h) = csc(x + h)^2

(f(x + h) - f(x)) / (x + h - x) =>

(csc(x + h)^2 - csc(x)^2) / h =>

(1/sin(x + h)^2 - 1/sin(x)^2) / h =>

((sin(x)^2 - sin(x + h)^2) / (sin(x + h)^2 * sin(x)^2)) / h =>

(sin(x)^2 - sin(x + h)^2) / (h * sin(x)^2 * sin(x + h)^2) =>

((1/2) * (1 - cos(2x)) - (1/2) * (1 - cos(2x + 2h))) / (h * sin(x)^2 * sin(x + h)^2) =>

(1/2) * (1 - cos(2x) - 1 + cos(2x + 2h)) / (h * sin(x)^2 * sin(x + h)^2) =>

(cos(2x + 2h) - cos(2x)) / (2h * sin(x)^2 * sin(x + h)^2) =>

(cos(2x)cos(2h) - sin(2x)sin(2h) - cos(2x)) / (2h * sin(x)^2 * sin(x + h)^2) =>

(cos(2x) * (cos(2h) - 1) - sin(2x) * 2sin(h)cos(h)) / (2h * sin(x)^2 * sin(x + h)^2) =>

(cos(2x) * (cos(h)^2 - sin(h)^2 - 1) - sin(2x) * 2sin(h)cos(h)) / (2h * sin(x)^2 * sin(x + h)^2) =>

(cos(2x) * (-sin(h)^2 - sin(h)^2) - sin(2x) * 2sin(h)cos(h)) / (2h * sin(x)^2 * sin(x + h)^2) >

(-2 * sin(h)^2 * cos(2x) - 2 * sin(2x) * sin(h) * cos(h)) / (2h * sin(x)^2 * sin(x + h)^2) =>

-2 * sin(h) * (sin(h) * cos(2x) + sin(2x)cos(h)) / (2h * sin(x)^2 * sin(x + h)^2) =>

-sin(h) * (sin(h) * cos(2x) + sin(2x)cos(h)) / (h * sin(x)^2 * sin(x + h)^2)

An important limit to remember is this: sin(h)/h goes to 1 as h goes to 0

-(sin(h)/h) * (sin(h) * cos(2x) + sin(2x) * cos(h)) / (sin(x)^2 * sin(x + h)^2)

Let h go to 0

-1 * (0 * cos(2x) + sin(2x) * 1) / (sin(x)^2 * sin(x)^2) =>

-1 * sin(2x) / sin(x)^4 =>

-2sin(x)cos(x) / sin(x)^4 =>

-2cos(x) / sin(x)^3 =>

-2 * cot(x) * csc(x)^2

Now, let's use some rules of differentiation to see if this checks out:

csc(x)^2

Derive

2 * csc(x) * (-cot(x) * csc(x)) =>

-2 * cot(x) * csc(x)^2

Looks right to me.

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- alexLv 74 weeks ago
Is that (cosec x)^2 or cosec (x^2) ?

- Mr. Rehan Baig4 weeks agoReport
cosec x^2

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