DLS
  • DLS
If "this" is true,then prove that (x^2-y^2+3)dy/dx=1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
|dw:1354339128951:dw|
anonymous
  • anonymous
is that x+1/n+1/x+1/x to infinity?
anonymous
  • anonymous
I see what you mean but now that I understand the question I would have to say that I am not sure either :/. Hopefully someone more knowledgeable than myself looks at this...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DLS
  • DLS
1 Attachment
DLS
  • DLS
Here's the printed question,just if its not clear
DLS
  • DLS
If "this" is true,then prove that (x^2-y^2+3)dy/dx=1
DLS
  • DLS
To prove: \[(x^{2}-y^{2}+3)\frac{dy}{dx}=1\]
DLS
  • DLS
@Algebraic! @ghazi @RadEn
ghazi
  • ghazi
use this \[Y=X+\frac{ 1 }{ Y }\]
sirm3d
  • sirm3d
the given equation \[\large y=x+\frac{ 1 }{ x+\frac{ 1 }{ x+... } }\] is equivalent to \[\large y=x+\frac{ 1 }{ y }\]
sirm3d
  • sirm3d
find the derivative by implicit differentiation
ghazi
  • ghazi
also you can substitute x at the place where you have to prove a desired result
DLS
  • DLS
please ellaborate,I'm doing this type of question first time
sirm3d
  • sirm3d
\[\large y = x+\frac{ 1 }{ \left[ x+\frac{ 1 }{ x+... } \right] }=x+\frac{ 1 }{ y }\] because the bracketed expression is equal to y
DLS
  • DLS
yeah i got that,what next
anonymous
  • anonymous
Ahh, that's clever sirm3d, makes sense now. I couldn't figure this one out.
sirm3d
  • sirm3d
just take the derivative of both sides by implicit differentiation. the derivative of the LHS is \[\large (dy/dx)\] the derivative of the RHS is 1 -
ghazi
  • ghazi
LHS --- left hand side RHS--- right hand side
DLS
  • DLS
lol i know that
anonymous
  • anonymous
Here is a good reference for implicit differentiation: http://www.intmath.com/differentiation/8-derivative-implicit-function.php
ghazi
  • ghazi
lol
sirm3d
  • sirm3d
oops, the derivative of the RHS is \[\large 1 - \frac{ 1 }{ y^2 }(dy/dx)\]
ghazi
  • ghazi
@DLS i guess you can do it now
DLS
  • DLS
what did u do with 1/y
sirm3d
  • sirm3d
\[\large \frac{ 1 }{ y }=y^{-1}\]
ghazi
  • ghazi
he differentiated it , implicitly
DLS
  • DLS
how did u get y^2
ghazi
  • ghazi
\[\frac{ dy (x^n) }{ dx }=nx^{n-1}\]
sirm3d
  • sirm3d
by power rule and chain rule, the derivative of y^(-1) is \[-1y^{-2} \frac{ dy }{ dx }\]
DLS
  • DLS
i know that too ! :o
DLS
  • DLS
okay whats the last step?
sirm3d
  • sirm3d
equate the derivative of the LHS to the derivative of the RHS, then group terms with (dy/dx)

Looking for something else?

Not the answer you are looking for? Search for more explanations.