If x^3 - 2x^2y^2 + 5x + y - 5 = 0, then at (1 | vedclass

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$(D)$ $	ext{Given the equation: } x^3 - 2x^2y^2 + 5x + y - 5 = 0$.
$	ext{Taking the first derivative with respect to } x:$
$3x^2 - (4xy^2 + 4x^2yy

Vedclass · Accepted Answer

$ -238/27 $

Vedclass · Answer

$(D)$ $	ext{Given the equation: } x^3 - 2x^2y^2 + 5x + y - 5 = 0$.
$	ext{Taking the first derivative with respect to } x:$
$3x^2 - (4xy^2 + 4x^2yy') + 5 + y' = 0$.
$	ext{At point } (1, 1), 	ext{ substitute } x=1 	ext{ and } y=1:$
$3 - (4 + 4y') + 5 + y' = 0$
$\Rightarrow 3 - 4 - 4y' + 5 + y' = 0$
$\Rightarrow 4 - 3y' = 0$
$\Rightarrow y' = \frac{4}{3}$.
$	ext{Now, differentiate the first derivative equation with respect to } x:$
$6x - [4y^2 + 8xyy' + 8xyy' + 4x^2(y')^2 + 4x^2yy''] + y'' = 0$.
$	ext{Substitute } x=1, y=1, y'=\frac{4}{3}:$
$6 - [4 + 8(4/3) + 8(4/3) + 4(16/9) + 4y''] + y'' = 0$
$6 - 4 - 64/3 - 64/9 - 4y'' + y'' = 0$
$2 - 192/9 - 64/9 - 3y'' = 0$
$2 - 256/9 = 3y''$
$(18 - 256)/9 = 3y''$
$-238/9 = 3y''$
$y'' = -238/27$.

If $x^3 - 2x^2y^2 + 5x + y - 5 = 0$, then at $(1, 1)$, $y''_1 = $?

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