¿Cómo se diferencian -2y = y ^ 2 / (xsin (x-y)?

¿Cómo se diferencian -2y = y ^ 2 / (xsin (x-y)?
Anonim

Responder:

# dy / dx = - (2sin (x-y) + 2xcos (x-y)) / (1-2xcos (x-y)) #

Explicación:

Podemos reorganizar y simplificar para obtener:

# -2xsin (x-y) = y #

# d / dx y = d / dx -2xsin (x-y) #

# d / dx y = d / dx -2x sin (x-y) -2xd / dx sin (x-y) #

# d / dx y = - 2sin (x-y) -2xd / dx sin (x-y) #

# d / dx y = - 2sin (x-y) -2xcos (x-y) d / dx x-y #

# d / dx y = - 2sin (x-y) -2xcos (x-y) (d / dx x -d / dx y) #

# d / dx y = - 2sin (x-y) -2xcos (x-y) (d / dx x -d / dx y) #

Usando la regla chqain obtenemos que # d / dx = dy / dx * d / dy #

# dy / dxd / dy y = - 2sin (x-y) -2xcos (x-y) (1-dy / dxd / dy y) #

# dy / dx = -2sin (x-y) -2xcos (x-y) (1-dy / dx) #

# dy / dx = -2sin (x-y) -2xcos (x-y) + 2xcos (x-y) dy / dx #

# dy / dx-2xcos (x-y) dy / dx = -2sin (x-y) -2xcos (x-y) #

# dy / dx 1-2xcos (x-y) = - 2sin (x-y) -2xcos (x-y) #

# dy / dx = - (2sin (x-y) + 2xcos (x-y)) / (1-2xcos (x-y)) #