`G(y)=(y-1)^4/(y^2+2y)^5`

Derivative can be found either by product rule or quotient rule.

`G(y)=(y-1)^4 (y^2+2y)^-5`

Using product rule

`G'(y)=(y-1)^4 d/(dy) (y^2+2y)^-5 + (y^2+2y)^-5 d/(dy) (y-1)^4`

`G'(y)=(y-1)^4 (-5(y^2+2y)^-6 (2y+2)) +(y^2+2y)^-5 (4(y-1)^3)`

`G'(y)=(-5(2y+2)(y-1)^4)/(y^2+2y)^6 + (4(y-1)^3)/(y^2+2y)^5`

`G'(y)=(-5(2y+2)(y-1)^4 + 4(y-1)^3(y^2+2y))/(y^2+2y)^6`

`G'(y)=((y-1)^3(-5(2y+2)(y-1) + 4y^2+8y))/(y^2+2y)^6`

`G'(y)=((y-1)^3(-10y^2+10y-10y+10+4y^2+8y))/(y^2+2y)^6`

`G'(y)=((y-1)^3(-6y^2+8y+10))/(y^2+2y)^6`

`G'(y)=(2(-3y^2+4y+5)(y-1)^3)/(y^2+2y)^6`

*Note:- 1) If y = x^n ; then dy/dx = n*{x^(n-1)}*

*2) If a function to be differentiated contains sub-functions,then by the rule of differentiation, the last function is differentiated first.*

*3) If the function is of the form y = u/v ; where u & v are both functions of 'x' , then dy/dx = y' = [{v*u' - u*v'}/(v^2)]*

*Now, for the given question , find the solution in the attachment*