If y = [(x+1)(2x+1)(3x+1) (nx+1)]^4 , where n N and dy dx at x = 0 is 2k , then the value of k is

Step-by-step solution

Given y = [(x+1)(2x+1)(3x+1) (nx+1)]^4 Taking natural logarithm on both sides: y = 4 [ (x+1) + (2x+1) + (3x+1) + + (nx+1)] Differentiating with respect to x : 1 y dy dx = 4 [ 1 x+1 + 2 2x+1 + 3 3x+1 + + n nx+1 ] At x = 0 , y = [(0+1)(0+1)(0+1) (0+1)]^4 = 1 Substituting x = 0 and y = 1 in the derivative expression: 1 1 ( dy dx )_ x=0 = 4 [1 + 2 + 3 + + n] ( dy dx )_ x=0 = 4 [ n(n+1) 2 ] = 2n(n+1) Given that ( dy dx )_ x=0 = 2k 2k = 2n(n+1) k = n(n+1) Answer: n(n+1)