Differentials of Trigonometric Functions

not sure where this belongs but

i am trying to diferentiate a function and it is y=cosec^2 x + sin2x^2

now using the product rule u = cosec^2 x and v = son2x^2

so dv/dx = 4xcos2x^2


can anybody aid me in this


Not at all, and thank you for making me ed hurt :haddock:

Errm duh…lol,
Sorry m8 absolutly not a clue…but out of curiosity… what is it?

edit… just to try and be helpfull… take a look at this… it may help…
Trigonometric function

cosec differentiates to -cosec x . cot x

Does it use the product rule? I thought product rule would be if y=cosec^2.sin2x^2? (. = multiply if you didn’t know…)

Believe it’s similiar to differentiating a polynominal one.

so dy/dx = 2(-cosec.cot) + 4cos?

Ebuyer sell new keyboards for £1.99, a couple of you seem to have defective ones

Could be the lead, sometimes that generates gibberish if faulty… :wink:

apologies it is CHAIN rule,

and thanks so cosec —> -cosec x . cot x

tyvm now i can finally finsih the math work


the answer is 42

Glad I could help.

Uni work is good for some things…

I am so glad I don’t have to do that any more…

To be honest, I’m going to have to do some math review to follow some of that. So much for the math skills.

lol, then anwer to everything is 42 lol


Deep Thought (the second greatest computer ever built) spent Millennia calculating that the ultimate answer to the ultimate question of Life, the Universe and Everything is ‘42’

As this is totally useless unless you know the actual question, Deep Thought designed the Earth (the greatest computer ever built) as a living computer to calculate that very question.

The ultimate question is ‘what do you get if you multiply 9 by 6’.


Q: What do you get if you multiply 9 by 6
A: 42

(and Mathematically it works!)


4xCOS2x^2 / cosec^2 ???

4 multiplied by Cos of 2 x to the power of 2 over the cosec squared…

yes… that was my answer… don’t know where it came from… perhaps this bottle of whiskey?!

i believe you split it into qoutient rule and product rule hence 4cos2x squared over cosec squared is equivalent to

u = 4cos2x squared and v = cosec squared

vdu by dx = cosec squared . -4sin2x

udv by dx = 4cos2x . now cosec = 1/sin and dy by dx sin = cos so

udv by dx = 4cos2x . (sin squared)^-1 (-1 gets it 1 over)

so you get vdu by dx - udv by dx (stated above)

divided by cosec squared squared…cosec to the four hence

dy/dx of the above =

(cosec squared . -4sin2x) - (4cos2x . (sin squared)^-1
Cosec to the power 4

ee walla

EDIT : lol just re read the post and i just did dy^2/dx so i differentated the answer lol, oh well it looks good so it stays

I can understand Differentials but only when filled with EP90

I’m with TFW on that one :whiteflag:

I did? no no no… I only have my grade 12 math :slight_smile: My girlfriend did that… :slight_smile: