Question

if 2 X square + 3 X Y + Y square + X + 2 Y - 8 is equals to zero then DY by DX is equal to

Answer 1

I HOPE THIS WILL HELP YOU

Answer 2

$Given \: 2x^{2} + 3xy + y^{2}+x+2y-8 = 0$

$\frac{d}{dx}(2x^{2} + 3xy + y^{2}+x+2y-8)= 0$

$\implies 2\times 2x + 3[ x\frac{dx}{dy}+y] + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0$

$\implies 4x + 3 x\frac{dx}{dy}+3y+ 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0$

$\implies \frac{dy}{dx}(3x+2y+2)= (-4x-3y-1)$

$\implies \frac{dy}{dx} = \frac{ - (4x+3y+1)}{(3x+2y+2)}$

Therefore.,

$\red{\frac{d}{dx}(2x^{2} + 3xy + y^{2}+x+2y-8)= 0}$

$\green { = \frac{ - (4x+3y+1)}{(3x+2y+2)}}$

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