Question

find y' if x^2y^4+3y-4x^3=5cos x-1
(2x*y^4)+dy/dx 4y^3*x^2)+dy/dx3+12x^2=5-sin-0
dy/dx=-5sin (x -12 x^2)/(2x*4y^3)+(12y^2+x^2)
this is using implicit differenation is this correct

Answer 1

d/dx of

x^2y^4+3y-4x^3=5cos x-1
((( note - dx/dx = 1)))
so
x^2[4y^3dy/dx] +y^4[2x]-12x^2 = -5sin x

4x^2y^3 dy/dx = 12x^2 -2xy^4-5sinx

dy/dx=3/y^3 - y/2x- 5sinx/(4x^2y^3)

Answer 2

(2x*y^4)+y'4y^3*x^2)+ 3y'+12x^2=-5sinx
On that line I am not certain if the original problems was 5cos(x-1) or 5cos x -1. I assume the latter. Then gathering terms..

y'(4x^2y^3+3)=(-5sinx -2xy^4-12x^2)

I do not see this leading to a y' which is the same as your manipulations.

Answer 3

x^2[4y^3dy/dx] +y^4[2x]+3 -12x^2 = -5sin x

4x^2y^3 dy/dx = 12x^2 -2xy^4-5sinx -3

dy/dx=3/y^3-y/2x-(3+5sinx)/(4x^2y^3)

Answer 4

x^2[4y^3dy/dx] +y^4[2x]+3dy/dx -12x^2 = -5sin x

[4x^2y^3+3]dy/dx = 12x^2 -2xy^4-5sinx

dy/dx
=(12x^2 -2xy^4-5sinx)/[4x^2y^3+3]