Part (1)
y=x+1x+2(1(x+2)2)
y=x+1(x+2)3
On differentiating both sides w.r.t x, we get
dydx=ddx[x+1(x+2)3]
dydx=⎡⎢ ⎢ ⎢⎣(x+2)3(1+0)−(x+1)(3(x+2)2(1+0))[(x+2)3]2⎤⎥ ⎥ ⎥⎦
dydx=(x+2)3−3(x+1)(x+2)2(x+2)6
dydx=(x+2)−3(x+1)(x+2)4
dydx=x+2−3x−3(x+2)4
dydx=−2x−1(x+2)4
dydx=−(2x+1)(x+2)4
Hence, this is the answer.
Part (2)
y=xex[(x+1)ex]
y=xe2x(x+1)
y=x2e2x+xe2x
On differentiating both sides w.r.t x, we get
dydx=ddx(x2e2x+xe2x)
dydx=x2e2x(2)+e2x(2x)+xe2x(2)+e2x(1)
dydx=2x2e2x+2xe2x+2xe2x+e2x
dydx=2x2e2x+4xe2x+e2x
dydx=e2x(2x2+4x+1)
Hence, this is the answer.