# 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 –y)2 Thanks a lot, this is for my last chance of...

Derivade of:

x^{3} + y^{2} + senx = cosy + y

If you can, please explain this others too:

- xy + x + y = 5 – x
- ctg
^{3}y + 2x – 3y = 8 - xy = (1 –x –y)
^{2}

Thanks a lot, this is for my last chance of pass calculus!

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### 2 Answers

x3 + y2 + sinx = cosy + y

(x^3)' + (y^2)' + (sinx)' = (cosy)' + y'

3x^2 + 2yy' + cosx = (siny)*y' + y'

Group terms with y; on the left sides:

==> (siny)y' + y' - 2yy' = 3x^2 + cosx

Now factor y':

==> y'(siny + 1 - 2y) = (3x^2 + cosx)

**==> y' = (3x^2 + cosx)/(siny -2y + 1)**

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).