Question

Consider the curve defines by 2y^3+6x^2y-12x^2+6y=1

Answer 1

\[2y^{3}+6x^{2}y-12x^{2}+6y=1\]

Answer 2

Show that \[\frac{ dy }{ dx }=\frac{ 4x-2xy }{ x^{2}+y^{2}+1 }\]

Answer 3

Do you know how to implicitly differentiate?

Answer 4

kinda, I put dy/dx when there is a y

Answer 5

\[2 y^3+6 x^2y-12 x^2+6y=1\]
diff. w.r.t. x
\[6 y^2\frac{ dy }{ dx }+6\left( 2xy+x^2\frac{ dy }{ dx } \right)-24x+6\frac{ dy }{ dx }=0\]
divide by 6 and then find dy/dx

Answer 6

How did you get 6(2xy+x2dydx)

Answer 7

Did you factor?

Answer 8

\[\frac{ d }{ dx }\left( uv \right)=u \frac{ dv }{ dx }+v \frac{ du }{ dx }\]

Answer 9

You used the product rule then factored?

Answer 10

keep 6 outside then differentiate \[x^2y\]

Answer 11

So after dividing I get: So after dividing I get: \[6y^2+2xy+x^{2}\frac{ dy }{ dx }-24x+6\frac{ dy }{ dx }=0\]

Answer 12

Subtract the 2xy now?

Answer 13

@campbell_st

Answer 14

Ok so I get \[x^2\frac{ dy }{ dx }+6\frac{ dy }{ dx }=-6y^2-2xy-24x\]

Answer 15

\[y^2\frac{ dy }{ dx }+2 xy+x^2\frac{ dy }{ dx }-4x+\frac{ dy }{ dx }=0\]
\[\left( y^2+x^2+1 \right)\frac{ dy }{ dx }=4x-2xy\]
\[\frac{ dy }{ dx }=?\]

Answer 16

Am I doing it right so far?

Answer 17

Where did the dy/dx next to the y^2 come from?

Answer 18

Ohh I see I forgot

Answer 19

Then divide everythign inside the parentheses to both sides

Answer 20

ohh I got it now hehe thanks ^_^

Answer 21

Now it asks me to write an equation of each horizontal tangent line to the curve, what exactly does that mean? I'm confused

Answer 22

put \[\frac{ dy }{ dx }=0\]