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  • DLS

If "this" is true,then prove that (x^2-y^2+3)dy/dx=1

Mathematics
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  • DLS
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is that x+1/n+1/x+1/x to infinity?
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...

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Other answers:

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

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