DLS
If "this" is true,then prove that (x^2-y^2+3)dy/dx=1
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DLS
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|dw:1354339128951:dw|
Kelumptus
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is that x+1/n+1/x+1/x to infinity?
Kelumptus
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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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DLS
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Here's the printed question,just if its not clear
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If "this" is true,then prove that (x^2-y^2+3)dy/dx=1
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To prove:
\[(x^{2}-y^{2}+3)\frac{dy}{dx}=1\]
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@Algebraic! @ghazi @RadEn
ghazi
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use this \[Y=X+\frac{ 1 }{ Y }\]
sirm3d
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the given equation \[\large y=x+\frac{ 1 }{ x+\frac{ 1 }{ x+... } }\] is equivalent to \[\large y=x+\frac{ 1 }{ y }\]
sirm3d
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find the derivative by implicit differentiation
ghazi
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also you can substitute x at the place where you have to prove a desired result
DLS
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please ellaborate,I'm doing this type of question first time
sirm3d
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\[\large y = x+\frac{ 1 }{ \left[ x+\frac{ 1 }{ x+... } \right] }=x+\frac{ 1 }{ y }\] because the bracketed expression is equal to y
DLS
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yeah i got that,what next
Kelumptus
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Ahh, that's clever sirm3d, makes sense now. I couldn't figure this one out.
sirm3d
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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
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LHS --- left hand side
RHS--- right hand side
DLS
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lol i know that
ghazi
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lol
sirm3d
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oops, the derivative of the RHS is \[\large 1 - \frac{ 1 }{ y^2 }(dy/dx)\]
ghazi
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@DLS i guess you can do it now
DLS
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what did u do with 1/y
sirm3d
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\[\large \frac{ 1 }{ y }=y^{-1}\]
ghazi
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he differentiated it , implicitly
DLS
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how did u get y^2
ghazi
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\[\frac{ dy (x^n) }{ dx }=nx^{n-1}\]
sirm3d
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by power rule and chain rule, the derivative of y^(-1) is \[-1y^{-2} \frac{ dy }{ dx }\]
DLS
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i know that too ! :o
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okay whats the last step?
sirm3d
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equate the derivative of the LHS to the derivative of the RHS, then group terms with (dy/dx)