Question

assume x and y are functions of t. Evaluate dy/dt for 2xy--4x/3y^3=48, with conditions dx/dt=-6,x=3, y=-2

Answer 1

Just use the product rule and chain rule:

2xy - 4x/3y^3 = 48

2y dx/dt + 2x dy/dt - 4/3y^3 dx/dt + 4x/y^4 dy/dt = 0

Now just plug in the numbers and solve for dy/dt

Answer 2

d/dt (2xy-4x/3y^3)= d/dt (48)
2*d(xy)/dt - d(4x/3y^3)/dt = 0
(because derrvation of constant is 0)
2* ((dx/dt)*y + xdy/dt) -(((d(4x)/dt)*3y^3^-4x*d(3y^3)/dt))/(3y^3)^2)=0
2*((-6)*(-2)+3*(dy/dt))-(4*(dx/dt)*(3y^3)-4*x*3*(3*y^2)*(dy/dt) )/9y^6 =0

2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) =0

2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) =0

24+6*(dy/dt)-((-24)*(-24)-36*(12)*(dy/dt) )/(9*64) =0

24+6*(dy/dt)-(576-432*(dy/dt) )/(576) =0
24+6*(dy/dt)-(576/576-432*(dy/dt)/576) =0
24+6*(dy/dt)-(1-3*(dy/dt)/4) =0
24+6*(dy/dt)-(1-3*(dy/dt)/4) =0 =0
23+(6+(3/4))*(dy/dt)=0
23+(24/4 + 3/4) *(dy/dt)=0
(27/4 ) *(dy/dt)= -23
(dy/dt)= (-23 ) /(27/4)
(dy/dt) = (-23/1 ) /(27/4)
(dy/dt) = (-23*4/27 )
(dy/dt) = (-92/27
du/dx=-3.41 roundet on 2-nd decimal place )

Answer 3

That's a lot of places to make mistakes. I get

2y dx/dt + 2x dy/dt - 4/3y^3 dx/dt + 4x/y^4 dy/dt
2y(1 - 2/3y^4) dx/dt + 2x(1+2/y^4) dy/dt = 0
-4(23/24)(-6) + 6(9/8) dy/dt = 0
dy/dt = 27/92

Maybe you can work out whether we are both wrong!

Actually, I plugged in the line

2*(12+3*(dy/dt))-(4*(-6)*(3*(-2)^3)-4*3*3*(3*(-2)^2)*(dy/dt) )/(9*(-2)^6) = 0

at wolframalpha and got dy/dx = -92/27

So, maybe you better do it yourself... Watch out for details.