Derivade of: x3 + y2 + senx = cosy + yIf you can, please explain this others too: xy + x + y = 5 – x ctg3y + 2x – 3y = 8 xy = (1 –x...

To find the derivatives dy/dx or
y':

1)xy+x+y = 5-x.Here find the derivatives term by term
and then group the terms with y' and solve for y'.

(xy)'
+x' +y' = (5-x)'.

x'y+xy' +1 + y' = (5)'-
x'.

y+xy' +1+ y' = 0-1.

xy' +
y' = -y-2.

(x+1)y' =
-(y+2).

y' =
(y+2)/(x+1).

2)ctg^3y +2x-3y =
8.

Differentiate with respect to
x:

(ctg^3y)' = {ctgy)^3} = 3ctg^2y*(ctgy)' =
3ctg^2y*(-cosec^2y)*y'=-3y'cos^2y.

(2x)' =
2.

(-3y)' = -3y'.

(8)' =
0.

Therefore (ctg^3y +2x-3y)' =
(8)'.

-3y'cosy +2-3y' = 0.

Now
we group the terms with y' together and solve for
y':

-3y'cosy -3y' =
-2.

-3y'(3+cosy) = -2.

y'  =
2/(3+cosy).

3)xy =
(1-x-y)^2.

Differentiate both
sides.

(xy)' = {(1-x-y)^2
}'.

xy'+y =
2(1-x-y)^(2-1)*(1-x-y)'.

xy'+y =
2(1-x-y)(-1-y').

xy' +y = -2+2x+2y  -
2y'(1-x-y).

xy'+2y'(1-x-y) =
-2+2x+2y-y.

y'(2-x-2y) =
-2+2x+y.

y' = (-2+2x+y)/(2-x-2y).