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(i)
Differentiating the given equation with respect to x,

$1 - \frac{dy}{dx} = 2(x+y)(1 + \frac{dy}{dx})$

$1 = 2x + 2y + (2x+2y+1)\frac{dy}{dx}$

$2 = (2x+2y+1)(1 + \frac{dy}{dx})$

$1 + \frac{dy}{dx} = \frac{2}{2x+2y+1}$

(ii)
$\frac{d^2y}{dx^2} = - \frac{2}{(2x+2y+1)^2} (2 + 2\frac{dy}{dx})$

$= - (\frac{2}{2x+2y+1})^2 (1 + \frac{dy}{dx})$

$= - (1+ \frac{dy}{dx})^3$

(iii)
When $latex \frac{dy}{dx} = 0, \frac{d^2y}{dx^2} = -1 <0$ Thus we conclude that the turning point is a maximum point.

### KS Comments:

Some students can still get the implicit differentiation wrong. (iii) actually is very simple if students do not overthink and trust in what they understand.

### One Comment

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