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Hidden_Twilight
 2 years ago
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?
Hidden_Twilight
 2 years ago
anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2  a^2)]?

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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2there's a dx there..so, do u mean to ntegrate ?

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1x = asecu dx = asecutanu sqrt(x^2  a^2) = atanu therefor:

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1372397190010:dw

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2hey thats absolutely correct! but don't forget we haven't integrated yet!

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2\(\huge \int a \: du=..?\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2oh, and one error, \(dx = a \sec u\ tanu \: du\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2not da, here the variable us 'u' , so 'du'

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1oh. so it becomes au?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2correct :) au +c now substitute back for 'u' x=a sec u so, u=...?

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1u = asec/x wait. does sec still have u or do i change it to x?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2u = asec/x ......no \(\large x = a \sec u \\ \large (x/a)=\sec u \\ \large u = \sec^{1}(x/a)\) the inverse secant function!

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1ahhhh... so it becomes a(sec^1 [x/a]) right?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2correct! don't forget the +c part :)

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1wait. does that mean i have to cancel a?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2you cannot cancel 'a' the other a is inside the inverse secant function

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2a(sec^1 [x/a]) +c will be the final answer

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1ohh...got it! thank you :)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2welcome ^_^ ask if anymore doubts...

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1btw. what if the other sie has y? a(sec^1 [x/a]) = y (y was already integrated) a(sec^1 [x/a]) = y + C (sec^1 [x/a]) =( y + C)/a ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2ohh...differential equations ? correct, go on... x/a = sec [(y+c)/a ] or x = a sec [(y+c)/a] ....

Hidden_Twilight
 2 years ago
Best ResponseYou've already chosen the best response.1ahhhh....so thats how it is. thanks har!! :D
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