Let y = (x+1)/(x+2)^2

Let y = u/v such that:

u= (x+1) ==> u' = 1

v= (x+2)^2 ==> v' = 2(x+2) = 2x + 4

==> y' = (u'v-uv')/v^2

= (x+2)^2 - (x+1)(2x+4)]/(x+2)^4

= (x^2 + 4x + 4 - 2x^2 -6 -4)/(x+2)^4

= (-x^2 -2x )/(x+2)^4

= -x(x+2)/(x+2)^4

= -x/(x+2)^3

**==> y' = -x/(x+2)^3**

We'll differentiate the first ratio, with respect to x, applying the quotient rule:

(f/g)' = (f'*g - f*g')/g^2

(x+1)/(x+2)^2 = f(x)/g(x)

We can write the differentiating process in 2 ways:

d/dx[(x+1)/(x+2)^2] = [(x+2)^2d/dx(x+1) - (x+1)d/dx)/(x+2)^2]/)/[(x+2)^2]^2

or

[(x+1)/(x+2)^2]' = {(x+1)'*(x+2)^2 - (x+1)*[(x+2)^2]'}/[(x+2)^2]^2

[(x+1)/(x+2)^2]' = [1*(x+2)^2 - 2(x+1)(x+2)*(x+2)']/(x+2)^4

[(x+1)/(x+2)^2]' = [(x+2)^2 - 2(x+1)(x+2)]/(x+2)^4

We'll factorize the numerator by (x+2):

[(x+1)/(x+2)^2]' = (x+2)*(x+2-2x-2)/(x+2)^4

We'll combine like terms and we'll simplify:

[(x+1)/(x+2)^2]' = -x/(x+2)^3

**d/dx[(x+1)/(x+2)^2] = -x/(x+2)^3**

Now, we'll differentiate the ratio (3x+4)/(4x+5) using again the quotient rule:

d/dx[(3x+4)/(4x+5)] = [(3x+4)'*(4x+5)-(3x+4)*(4x+5)']/(4x+5)^2

d/dx[(3x+4)/(4x+5)] = [3(4x+5) - 4(3x+4)]/(4x+5)^2

We'll remove the brackets:

d/dx[(3x+4)/(4x+5)] =(12x+15-12x-16)/(4x+5)^2

We'll eliminate like terms:

**d/dx[(3x+4)/(4x+5)] = -1/(4x+5)^2**

To diffefrentiate (x+1)/(x+1)^2 .

We know that (x+1)/(x+1)^2 = 1/(x+1) .We divided both numerator and denominator by (x+1)

Therefore d/(dx(1(x+1)) = -1/(x+1)^2 , as d/dx(x+a)^n) = n(x+^(n-1).

To differentiate (3x+4)/(4x+5).

We simplify (3x+4)/(4x+5 ) =3(4x+5)/4(4x+5) - (4-15/4)/(x+5) = (3/4) - 1/4(4x+5)

Therefore {(1/4) (-1/(4x+5)^2) *(4x+5)' = -4/4(x+5)^2 = -1/(4x+5)^2 , as {(ax+b)^n }'= n(ax+b)^(n-1)* (ax+b)' = n(ax+b)^(n-1) * a

Therefore ,{(3x+4)/(4x+5)}' = --1/(4x+5)^2.