Question

HELP! ASAP! MEDAL!

Answer 1

@dan815 @freckles @Nnesha @satellite73

Answer 2

@satellite73 @zepdrix

Answer 3

have you differentiated yet?

Answer 4

no. I'm stuck at that part.

Answer 5

we need product rule and chain rule mainly

Answer 6

would the left-hand side be sin2y + x(cos 2y)?

Answer 7

\[\frac{d}{dx}(x \sin(2y)) =\sin(2y)\frac{d}{dx}x+x \frac{d}{dx}\sin(2y) \\ \frac{d}{dx}(x \sin(2y))=\sin(2y)+x \cdot 2y'\cos(2y)\]

Answer 8

we need chain rule for the sin(2y) part

Answer 9

derivative of inside * outside

Answer 10

(sin(2y))'
=
(2y)'*cos(2y)
=
2y'*cos(2y)

Answer 11

\[(x \sin(2y))'=(y \cos(2x))' \\ \sin(2y)+2y' x \cos(2y)=(y \cos(2x))'\]

Answer 12

the right hand side should be a bit easier now

Answer 13

I'm confused....How'd you get (y cos (2x))'?

Answer 14

I think you equation is
x sin(2y)=y cos(2x) right?

Answer 15

Oh, right sorry. you put it together. I thought you were saying the derivative of the left side was that, lol

Answer 16

cool looking
http://www.wolframalpha.com/input/?i=x*sin%282y%29%3Dy*cos%282x%29

Answer 17

Thanks. I have to go wat dinner. Hopefully I can figure out the rest myself. :/

Answer 18

eat*

Answer 19

k well put what you get for the right hand side
and I will check it

Answer 20

when you get time

Answer 21

I got \(\Large\frac{dy}{dx} = \frac{-y\cos{2x}-\sin{2x}}{-\sin{2y}+\cos{2y}}\)

Answer 22

\[(x \sin(2y))'=(y \cos(2x))' \\ \sin(2y)+2y' x \cos(2y)=y' \cos(2x)-2 y \sin(2x) \\ y'(2 x \cos(2y)-\cos(2x))=-2y \sin(2x)-\sin(2y) \\ \]
you dy.dx looks a little off

Answer 23

I changed it and ended up getting the equation y = 2x for my answer

Answer 24

(equation of line tangent to....)