If y=[(x+1)(2 x+1)(3 x+1) ( n x+1)]^2, then d y ~d x at x=0 is

Step-by-step solution

y=[(x+1)(2 x+1)(3 x+1) ( n x+1)]^2 Taking 'log' on both sides, we get aligned y=2[ (x+1)+ (2 x & +1)+ (3 x+1) & + + (n x+1)] aligned Differentiating w.r.t. x , we get aligned & 1 y ~d y ~d x =2 ( 1 x+1 + 2 2 x+1 + 3 3 x+1 + + n n x+1 ) & d y ~d x =2 y ( 1 x+1 + 2 2 x+1 + 3 3 x+1 + + n n x+1 ) aligned Now at x=0, y=[(1)(1)(1) (1)]^2=1 aligned ( d y ~d x )_ x=0 & =2(1) ( 1 0+1 + 2 0+1 + 3 0+1 + + n 0+1 ) & =2(1+2+3+ + n ) & =2 n ( n +1) 2 = n ( n +1) aligned