how do i solve this equation x+y=arctan y , to obtain y' ?

Answer 1

differentiate the both side!
1+y'=1/(1+y^2)y'
y'y^2/(1+y^2)=-1
y'=-1-1/y^2

Answer 2

x + y = arctan(y)

You have to differentiate implicitly.

1 + y' = [1/(1 + y^2)] y'

Bring everything with a y' in it to the left hand side; everything else goes to the right hand side.

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

Factoring y' out of the left hand side, we get

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

Put the stuff in the brackets under a common denominator.

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

Simplify the numerator.

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

Now, multiply both sides by the recipriocal of y^2 / (1 + y^2).

y' = (-1) (1 + y^2)/(y^2)

Answer 3

I think you're asking for y', the derivative?
I suggest implicit differentiation:

1 + dy/dx = (1/(1+y^2))(dy/dx)
whence
y' (1/(1+y^2) - 1) = 1
and
y' = -(1+y^2)/y^2
= -1/(y^2) - 1

Alternatively, you could make it
tan(x+y) = y

Hence (sec(x+y))^2 (1 + y') = y'
and so on, but I think the first way is better.

Answer 4

x + y = arctan(y)

tan(x + y) = y

differentiate both side:

( 1 + dy/dx)/cos(x + y)^2 = dy/dx

1 + y' = cos(x + y)^2y'

y'(1 - cos(x + y)^2) + 1 = 0

y' = -1/sin(x + y)^2

Answer 5

You need to differentiate implicitly.

x + y = arctan y

1 + dy/dx = {1/(1 + y²)}(dy/dx)
1 = (dy/dx){1/(1 + y²) - 1} = (dy/dx){(1 - 1 - y²)/(1 + y²)}
1 = (dy/dx){-y²/(1 + y²)}
dy/dx = 1/{-y²/(1 + y²)} = (1 + y²)/(-y²) = -(1 + y²)/y²

Answer 6

You cant. Whats x? You have 2 variables here. Do you want this in function of x? Add details, please

Ana