If y = x + x + x + , then ( dy dx )^2 at x = 4 is

Step-by-step solution

Given y = x + x + x + Squaring both sides, we get: y^2 = x + y Differentiating both sides with respect to x : 2y dy dx = ^2 x + dy dx dy dx (2y - 1) = ^2 x dy dx = ^2 x 2y - 1 At x = 4 , x = 1 . Substituting this into the equation for y : y^2 - y - 1 = 0 Solving for y , we get y = 1 5 2 . Since y represents a principal square root, y > 0 , so y = 1 + 5 2 . Substituting y = 1 + 5 2 and x = 4 into the derivative: 2y - 1 = 2 ( 1 + 5 2 ) - 1 = 5 ^2 ( 4 ) = ( 2 )^2 = 2 dy dx = 2 5 Squaring both sides: ( dy dx )^2 = 4 5