Find dy/dx by implicit differentiation.
(3xy^2 + 1)^4 = 2x - 3y
4(3xy^2 + 1)^3 (3xy^2 + 1) = 2 -3
(3xy^2 + 1) +3 = 2-4(3xy^2 +1)^3
dy/dx = 2 - 4(3xy^2 +1)^3/(3xy^2 +1) +3
Am I getting any closer to understanding this?
You started out ok, but your second line is incorrect.
This part is right 4(3xy^2+1)^3. Now you must multiply by the derivative of the inside function.
So d/dx (3xy^2+1)= 6xy(dy/dx)+3y^2.
Putting the LHS together it's 4(3xy^2+1)^3 * (6xy(dy/dx)+3y^2)
The RHS of your work is correct.
Hello, Becky!
I don't see dy/dx or y' in your work.
Then it suddenly appears in your last step.
Find dy/dx by implicit differentiation: .(3xy² + 1)^4 .= .2x - 3y
. . . . . . 4(3xy² + 1)³(6xyy' + 3y²) . = . 2 - 3y'
. . . . . . 24xy(3xy² + 1)3y' + 12y²(3xy² + 1)³ . = . 2 - 3y'
. . . . . . 24xy(3xy² + 1)3y' + 3y' . = . 2 - 12y²(3xy² + 1)³
Factor: .3[8xy(3xy(3xy² + 1)³ + 1]y' . = . 2[1 - 6y²(3xy² + 1)³]
. . . . . . . . . . . . . . .2[1 - 6y²(3xy² + 1)³]
Therefore: . y' . = . ---------------------------
. . . . . . . . . . . . . . 3[8xy(3xy² + 1)³ + 1]