How do you find #(dy)/(dx)# given #3y^3+2=2x#?

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Answer 1 · 2017-02-22T23:58:47

#dy/dt=2/(9y^2)# or #dy/dt=2/(9(((2/3)x-(2/3))^(1/3))^2)#

#9y^2(dy/dx)=2#

Solve for #dy/dt#

#dy/dt=2/(9y^2)#

If you want, we can solve and plug in for y using the given equation and simplifying:

#3y^3+2=2x#

#3y^3=2x-2#

#y^3=(2/3)x-(2/3)#

#y=((2/3)x-(2/3))^(1/3)#

Now plug this in for y and simplify.

#dy/dt=2/(9(((2/3)x-(2/3))^(1/3))^2)#

#dy/dt=2/(9(((2/3)x-(2/3))^(2/3)))#