Chapter 3.5, Problem 46E

x2⁄3 + y2⁄3

Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. π 4 (1, 0) Step 1 To find the equation of the tangent line to the graph of arctan(x + y) = y² + differentiate arctan(x + y) = y² + with respect to x. Therefore, -I 4 π Step 2 arctan(x + y) = y² + Submit d 1 1 + x + y 1 + 1 (arctan(x + y)) 1 [1 + y'] 1 To find the value of y' at the given point (1, 0), substitute x = 1 and y = 0 in + 0)² [1 = Skin (you cannot come back) = 2yy'. [1 + y'] = 20 -[1 + y' ] = = + ⇒ y' = 11) y' 1 1 + (x + y)² -[1 + y'] = 2yy' and solve for y'. Therefore,

1. Answer and give the complete solution: