# 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 –y)2 Thanks a lot!

hala718 | High School Teacher | (Level 1) Educator Emeritus

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xy + x + y = 5 – x

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

==> x'y + xy' + 1 + y1 = 0 - 1

==> y + xy' + 1 + y' = -1

==> xy' + y' = -1 -1 - y

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

==> y' = (-2-y)/(x+1)

x3 + y2 + sinx = cosy + y

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

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

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

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

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.