`arctan(x+y) = y^2 + pi/4 , (1,0)` Use implicit differentiation to find an equation of the tangent line at the given point

First, check that the given point satisfies the equation: `arctan(1 + 0) = 0 + pi/4` is true.

The slope of the tangent line is `y'(x)` at the given point. Differentiate the equation with respect to `x:`

`(1 + y')/(1+(x+y)^2) = 2yy'.`

Substitute `x = 1` and `y = 0` and obtain `1 + y' = 0,` so `y' = -1.`

Then the equation of the tangent line is  `y = -(x - 1) =...

First, check that the given point satisfies the equation: `arctan(1 + 0) = 0 + pi/4` is true.

The slope of the tangent line is `y'(x)` at the given point. Differentiate the equation with respect to `x:`

`(1 + y')/(1+(x+y)^2) = 2yy'.`

Substitute `x = 1` and `y = 0` and obtain `1 + y' = 0,` so `y' = -1.`

Then the equation of the tangent line is  `y = -(x - 1) = -x + 1.`