Derivade of: xy + x + y = 5 – xIf you can, please help me with this others: x3 + y2 + senx = cosy + y ctg3y + 2x – 3y = 8 xy = (1 –x...

1) For the first expression, we'll isolate the terms in y
to the left side, and we'll calculate the first derivative with respect to
x:

xy + x + y = 5 – x

xy + y =
5 - x - x

xy + y = 5 -
2x

We'll factorize by y to the left side and we'll
get:

y(x+1) = 5 - 2x

We'll
divide by (x+1) both sides and we'll have:

y =
(5-2x)/(x+1)

Now, we'll calculate the
derivative:

dy/dx =
(d/dx)[(5-2x)/(x+1)]

Since we have to calculate the
derivative of a quotient, we'll use the quotient
rule:

(f/g)' =
(f'*g-f*g')/g^2

We'll put f = 5-2x and g =
x+1

f' = -2

g' =
1

[(5-2x)/(x+1)]' = (-2(x+1) - 5 +
2x)/(x+1)^2

We'll remove the brackets from the right
side:

[(5-2x)/(x+1)]' = (-2x - 2 - 5 +
2x)/(x+1)^2

We'll combine and eliminate like
terms:

[(5-2x)/(x+1)]' =
-7/(x+1)^2

So, y' =
-7/(x+1)^2

2) x^3 + y^2 + sin x 
= cos y + y

We'll isolate theterms in x to the left side
and the terms in y to the right side.

We'll calculate the
derivative to the left side, with respect to x.

x^3 + sin x
= cos y + y - y^2

(d/dx)(x^3 + sin x) = 3x^2 + cos
x

We'll calculate the derivative to the right side, with
respect to y.

(d/dy)(cos y + y - y^2) = -sin y + 1 -
2y

The derivative of the given expression
is:

3x^2 + cos x = 1 - 2y - sin
y

3) (ctg y)^3 + 2x – 3y =
8

We'll keep the terms in y to the left side
and we'll move the terms in x to the right side:

(ctg y)^3
– 3y = 8 - 2x

We'll calculate the partial derivative, with
respect to y, to the left side:

 (d/dy)((ctg y)^3 – 3y) =
-3(ctg y)^2/(sin y)^2 - 3

We'll calculate the partial
derivative, with respect to x, to the right side:

 (d/dx)(8
- 2x) = 0 - 2

 (d/dx)(8 - 2x) =
-2

The derivative of the given expression
is:

-3(ctg y)^2/(sin y)^2 - 3 =
-2

(ctg y)^2/(sin y)^2 =
-1/3

 

4)
xy = (1 –x –y)^2

We'll square raise the right
side:

(1 –x –y)^2 = 1 + x^2 + y^2 - 2x - 2y +
2xy

The expression will
become:

xy = 1 + x^2 + y^2 - 2x - 2y +
2xy

1 + x^2 + y^2 - 2x - 2y + xy =
0

If we'll calculate the derivative with respect to x,
we'll get:

2x +  y^2 - 2 - 2y + y =
0

(d/dx)(1 + x^2 + y^2 - 2x - 2y + xy)' =  2x
+  y^2 - y - 2

Note: the terms in y are
considered constants.

If we'll calculate the derivative
with respect to y, we'll get:

(d/dy)(1 + x^2 + y^2 - 2x -
2y + xy)' = x^2 + 2y - 2x - 2 + x

(d/dy)(1 +
x^2 + y^2 - 2x - 2y + xy)' = x^2 + 2y - x -
2

Note: the terms in x are considered
constants.