DIFERENCIACION IMPLICITA

 x³ + y³ = 4  Se derivan ambos lados de la ecuaciòn

D ( x³ + y³ ) = D ( 4 ) ,

D ( x³ ) + D ( y³ ) = D ( 4 ) ,

3x² + 3y² y' = 0 ,

3y² y' = - 3x² ,

$y' = \displaystyle{ - 3x^2 \over 3y^2 } = \displaystyle{ - x^2 \over y^2 }$ .

 (x-y)² = x + y - 1

D (x-y)² = D ( x + y - 1 ) ,

D (x-y)² = D ( x ) + D ( y ) - D ( 1 ) ,

$2 (x-y) \ D (x-y) = 1 + y' - 0$ ,

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

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

- 2 (x-y) y' - y' = 1 - 2 (x-y) ,

(Factorizar respecto a  y' .)

y' [ - 2 (x-y) - 1 ] = 1 - 2 (x-y) ,

$y' = \displaystyle{ 1 - 2 (x-y) \over - 2 (x-y) - 1 } = \displaystyle{ 2y - 2x + 1 \over 2y - 2x - 1 }$ .

 $y = \sin(3x + 4y)$

$D(y) = D ( \sin(3x + 4y) )$ ,

$y' = \cos(3x + 4y) \ D ( 3x + 4y )$ ,

$y' = \cos(3x + 4y) ( 3 + 4 y' )$ ,

$y' = 3 \cos(3x + 4y) + 4 y' \cos(3x + 4y)$ ,

$y' - 4 y' \cos(3x + 4y) = 3 \cos(3x + 4y)$ ,

(Factorizando respecto a  y' .)

$y' [ 1- 4 \cos(3x + 4y) ] = 3 \cos(3x + 4y)$ ,

$y' = \displaystyle{ 3 \cos(3x + 4y) \over 1- 4 \cos(3x + 4y) }$ .

 y = x² y³ + x³ y²

D(y) = D ( x² y³ + x³ y² ) ,

D(y) = D ( x² y³ ) + D ( x³ y² ) ,

(Usando la regla del producto)

$y' = \{ x^2 D ( y^3 ) + D ( x^2 ) y^3 \} + \{ x^3 D ( y^2 ) + D ( x^3 ) y^2 \}$ ,

$y' = \{ x^2 ( 3y^2 y' ) + ( 2x ) y^3 \} + \{ x^3 ( 2 y y' ) + ( 3x^2 ) y^2 \}$ ,

y' = 3x² y² y' + 2x y³ + 2x³ y y' + 3x² y² ,

agrupar los términos que tienen y' en un lado de la ecuación

y' - 3x² y² y' - 2x³ y y' = 2x y³ + 3x² y² ,

(Factorizar respecto a y' .)

y' [ 1 - 3x² y² - 2x³ y ] = 2x y³ + 3x² y² ,

$y' = \displaystyle{ 2x y^3 + 3x^2 y^2 \over 1 - 3x^2 y^2 - 2x^3 y }$ .

 e^xy = e^4x - e^5y

D(e^xy ) = D ( e^4x - e^5y ) ,

D( e^xy ) = D ( e^4x ) + D ( e^5y ) ,

e^xy D( xy ) = e^4x D ( 4x ) + e^5y D( 5y ) ,

e^xy ( xy' + (1) y ) = e^4x ( 4 ) + e^5y ( 5y' ) ,

xe^xy y' + y e^xy = 4 e^4x + 5e^5y y' ,

xe^xy y' - 5e^5y y' = 4 e^4x - y e^xy ,

y' [ xe^xy - 5e^5y ] = 4 e^4x - y e^xy ,

$y' = \displaystyle{ 4 e^{4x} - y e^{xy} \over xe^{xy} - 5e^{5y} }$ .

 $\cos^2 x + \cos^2 y = \cos( 2x + 2y )$

$D( \cos^2 x + \cos^2 y ) = D ( \cos( 2x + 2y ) )$ ,

$D( \cos^2 x ) + D ( \cos^2 y ) = D ( \cos( 2x + 2y ) )$ ,

$(2 \cos x) D ( \cos x ) + (2 \cos y) D ( \cos y ) = - \sin( 2x + 2y ) D ( 2x + 2y )$ ,

$2 \cos x ( - \sin x ) + 2 \cos y ( - \sin y ) ( y' ) = - \sin( 2x + 2y ) ( 2 + 2y' )$ ,

$- 2 \cos x \sin x - 2 y' \cos y \sin y = - 2 \sin( 2x + 2y) - 2 y' \sin( 2x + 2y)$ ,

$2 y' \sin( 2x + 2y) - 2 y' \cos y \sin y = - 2 \sin( 2x + 2y) + 2 \cos x \sin x$ ,

$y' [ 2 \sin( 2x + 2y) - 2 \cos y \sin y ] = 2 \cos x \sin x - 2 \sin( 2x + 2y)$ ,

$y' = \displaystyle{ 2 \cos x \sin x - 2 \sin( 2x + 2y) \over 2 \sin( 2x + 2y) - 2 \cos y \sin y }$ ,

$y' = \displaystyle{ 2 [ \cos x \sin x - \sin( 2x + 2y) ] \over 2 [ \sin( 2x + 2y) - \cos y \sin y ] }$ ,

$y' = \displaystyle{ \cos x \sin x - \sin( 2x + 2y) \over \sin( 2x + 2y) - \cos y \sin y }$ .

 $x = \sqrt{ x^2 + y^2 }$

$D( x ) = D ( \sqrt{ x^2 + y^2 } )$ ,

1 = (1/2)( x² + y² )^-1/2 D ( x² + y² ) ,

1 = (1/2)( x² + y² )^-1/2 ( 2x + 2y y' ) ,

$1 = \displaystyle{ (1/2) (2) ( x + y y' ) \over \sqrt{x^2 + y^2} }$ ,

$1 = \displaystyle{ x + y y' \over \sqrt{x^2 + y^2} }$ ,

$\sqrt{x^2 + y^2} = x + y y'$ ,

$\sqrt{x^2 + y^2} - x = y y'$ ,

$y' = \displaystyle{ \sqrt{x^2 + y^2} - x \over y }$ .

 $${ x - y^3 \over y + x^2 } = x + 2$$ . multiplicando ambos lados  por  y + x²

$${ x - y^3 \over y + x^2 } (y + x^2 ) = (x + 2) (y + x^2 )$$ ,

x - y³ = xy + 2y + x³ + 2x² .

D ( x - y³ ) = D ( xy + 2y + x³ + 2x² ) ,

D ( x ) - D (y³ ) = D ( xy ) + D ( 2y ) + D ( x³ ) + D ( 2x² ) ,

1 - 3 y² y' = ( xy' + (1)y ) + 2 y' + 3x² + 4x ,

1 - y - 3x² - 4x = 3 y² y' + xy' + 2 y' ,

1 - y - 3x² - 4x = (3y² + x + 2) y' ,

$y' = \displaystyle{ 1 - y - 3x^2 - 4x \over 3y^2 + x + 2 }$ .

 $${ { y \over x^3 } + { x \over y^3 } } = x^2y^4$$ multiplicando ambos lados  por x³ y³

$\Big\{ \displaystyle{ { y \over x^3 } + { x \over y^3 } } \Big\} ( x^3 y^3 ) = x^2 y^4 ( x^3 y^3 )$ ,

$${ { y x^3 y^3 \over x^3 } + { x x^3 y^3 \over y^3 } } = x^2 x^3y^4 y^3$$ ,

y⁴ + x⁴ = x⁵ y⁷ .

D ( y⁴ + x⁴ ) = D ( x⁵ y⁷ ) ,

D ( y⁴ ) + D ( x⁴ ) = x⁵ D (y⁷ ) + D ( x⁵ ) y⁷ ,

4 y³ y' + 4 x³ = x⁵ (7 y⁶ y' ) + ( 5 x⁴ ) y⁷ ,

4 y³ y' - 7 x⁵ y⁶ y' = 5 x⁴ y⁷ - 4 x³ ,

y' [ 4 y³ - 7 x⁵ y⁶ ] = 5 x⁴ y⁷ - 4 x³ ,

$y' = \displaystyle{ 5 x^4 y^7 - 4 x^3 \over 4 y^3 - 7 x^5 y^6 }$ .

 (x²+y²)³ = 8x²y²

D (x²+y²)³ = D ( 8x²y² ) ,

3 (x²+y²)² D (x²+y²) = 8x² D (y² ) + D ( 8x² ) y² ,

3 (x²+y²)² ( 2x + 2 y y' ) = 8x² (2 y y' ) + ( 16 x ) y² ,

6x (x²+y²)² + 6 y (x²+y²)² y' = 16 x² y y' + 16 x y² ,

6 y (x²+y²)² y' - 16 x² y y' = 16 x y² - 6x (x²+y²)² ,

y' [ 6 y (x²+y²)² - 16 x² y ] = 16 x y² - 6x (x²+y²)² ,

$y' = \displaystyle{ 16 x y^2 - 6x (x^2+y^2)^2 \over 6 y (x^2+y^2)^2 - 16 x^2 y }$ .

 x²y + y⁴ = 4 + 2x

D ( x² y + y⁴ ) = D ( 4 + 2x ) ,

D ( x² y ) + D (y⁴ ) = D ( 4 ) + D ( 2x ) ,

( x² y' + (2x) y ) + 4 y³ y' = 0 + 2 ,

x² y' + 4 y³ y' = 2 - 2x y ,

y' [ x² + 4 y³ ] = 2 - 2x y ,

$y' = \displaystyle{ 2 - 2x y \over x^2 + 4 y^3 }$ .