Q1.
Let y = x−1(x+1)2
So y = x−1(x−1)2+4x=x−1(x−1)2+4(x−1)+4
=> y = (x-1) +4 + x−14
=> y = x + 3 + x−14
So dxdy=1+0−(x−1)24
Differentiating again
dx2d2y=0−(x−1)34∗(−2)=(x−1)38
Therefore D²(x−1(x+1)2 ) = (x−1)38
Q2. D(ln(x²+2x+1))
= D(ln(x+1)²)
= D(2ln(x+1)) since ln(A)n = n ln(A)
= 2D(ln(x+1))
= 2dxd(ln(x+1))
= x+12
Note: Given answer of question number 2 is wrong.
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