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Sergeu [11.5K]
3 years ago
13

Help please​ I will give brainliest

Mathematics
1 answer:
Helga [31]3 years ago
4 0

Answer:

B is the answer I believe.

Step-by-step explanation:

I think B

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  \boxed{18}

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Lina20 [59]
\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
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\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y


y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
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\\
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\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}

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3 years ago
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