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maks197457 [2]
2 years ago
11

AAAAA

Mathematics
1 answer:
mr_godi [17]2 years ago
7 0

Answer:

600

Step-by-step explanation:

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\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
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\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
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\\\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
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\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right)  (xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  (xy+1)^2 = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\2y\ y' + \left(3y+3x\ y'\right)  ((xy)^2+2xy+1) = 0
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\\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
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
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\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
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\\y' = \frac{3}{( -32 )}

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3 years ago
What are the vertical asymptotes of the function f(x) =5x+5/x2 + x-2
katen-ka-za [31]

let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.

\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}

5 0
2 years ago
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Solnce55 [7]

Answer:

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Step-by-step explanation:

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