Question

Find an equation of the tangent line to the graph of the equation `tan^-1 (x+y) = x^2 + pi/4`?

i need help in this

Answer 1

For any point, `(x_1, y_1)` on the graph, the tangent line is:

`y = (2x_1((x_1+y_1)^2+1)-1)(x-x_1)+ y_1`

Explanation:

The equation of the tangent line for any point `(x_1,y_1)` is:

`y = y'(x_1,y_1)(x-x_1)+ y_1" [1]"`

We need to compute `y'(x,y)` by implicit differentiation.

`(d(tan^-1 (x+y)))/dx = (d(x^2 + pi/4))/dx`

The right side is trivial

`(d(tan^-1 (x+y)))/dx = 2x`

The left side requires the use of the chain rule with `u = x+y`

`(d(tan^-1 (u)))/(du)(du)/dx = 2x`

We know that `(d(tan^-1 (u)))/(du) = 1/(u^2+1)`:

`1/(u^2+1)(du)/dx = 2x`

Multiply both sides by `u^2+1`

`(du)/dx = 2x(u^2+1)`

Substitute `u = x+y`:

`(d(x+y))/dx = 2x((x+y)^2+1)`

The last one is trivial:

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

Subtract 1 from both sides:

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

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

Substituting into equation [1]:

`y = (2x_1((x_1+y_1)^2+1)-1)(x-x_1)+ y_1`