**Moderator's edit**:

Given that \( x = y + \cfrac 1{y + \cfrac1{y + \cfrac1{y + \cfrac1{y + _\ddots}}}} \).

Show that \( \dfrac{dy}{dx} = 2x^2 + y^2 - 3xy \).

**Moderator's edit**:

Given that \( x = y + \cfrac 1{y + \cfrac1{y + \cfrac1{y + \cfrac1{y + _\ddots}}}} \).

Show that \( \dfrac{dy}{dx} = 2x^2 + y^2 - 3xy \).

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## Comments

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TopNewest\[x=y+\dfrac 1x \implies 1=x(x-y)\cdots (2)\\ \implies x^2=xy+1\cdots (1)\] Differentiating \((1)\) \[y'=(\dfrac{2x-y}{x})(1)\] Using \((2)\): \[y'=(\dfrac{2x-y}{\not x})(\not x(x-y))\] \[=2x^2-3xy+y^2\] – Rishabh Cool · 9 months ago

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– Akhilesh Prasad · 9 months ago

I think the minute i opened the note and saw the moderator's edit, that was when the answer came to me. Kudos to the moderator.Log in to reply

– Akhilesh Prasad · 9 months ago

Thanks a lot.Log in to reply