Calculate Y' and Y'' at the point (2,6)?

Question

On the curve $x^{2} + y^{2} = 4 x + 9 y - 22$ $x^2+y^2=4x+9y-22$

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Answer 1 · 2018-06-18T21:20:30

$y'=0$ , $y''=-2/3$

Differentiating the given expression implicitly wrt $x$
$2x+2ydy/dx=4+9dy/dx+0$ , i.e, $dy/dx[2y-9]=[4-2x]$

$y'=dy/dx=[4-2x]/[2y-9]$ and substituting for the given values of $y$ and $x$
$dy/dx=[4-2[2]]/[2[6]-9]$ = $0/3 = 0$ i.e, [ $y'=0]$

For the second derivative, $y''$ = $[d^2y]/dx^2$ = $d/dx[[4-2x]/[2y-9]]$

Using the quotient rule to differentiate this expression ie, $d/dx[u/v]=[vdu/dx-udv/dx]/[v^2$ where $u$ and $v$ are both functions of $x$ we obtain,

$[[[2y-9][-2] - [4-2x][2]]dydx]/[2y-9]^2$ , but we know that $dy/dx=0$ at $[2, 6]$

therefore, $y''= -2[2y-9]/[2y-9]^2$ = $-2/3$ when $y=6$ .