Derivative ?

Question

find the derivative of

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

Answer 1

Using quotient rule:-
G `(y) = N / D where :-
N =
(y² + 2y)(^5) (4)(y - 1)³ - (y - 1)(^4) (5)(y² + 2y)^(4)(2y + 2)

N =
(y² + 2y)^(4)(y - 1)³ [4 (y² + 2y) - (y - 1)(5)(2y + 2)]

N = (y² + 2y)^(4)(y - 1)³ [4y² + 8y - 5 (2y² - 2)]
N = (y² + 2y)^(4) (y - 1)³ [ 10 + 8y - 6y² ]

D = (y² + 2y)^(10)

N/D = (y - 1)³ (10 + 8y - 6y² ) / (y² + 2y)^(6)

Difficult to type---apologies in advance for any errors!

Answer 2

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

Answer 3

When it's a fraction, you take:
(derivative of top x bottom)-(derivative bottom x top) all divided by the bottom squared.

So....

{[4(y-1)^3(y^2+2y)^5]-[5(y^2+2y)^4 (2y+2) (y-1)^4]}/(y^2+2y)^10

Answer 4

To get this derivative, you need to know two things:
1) How to differentiate a product (same as a quotient); and
2) How to differentiate an expression to a power.

Differentiating the product Hi times Ho is:
Hi x d(Ho) + Ho x d(Hi)

Differentiating f(x)^n is:
n x f'(x) x f(x)^(n-1)

First, Hi=(y-1)^4 and Ho= (y^2+2y)^-5 (note the minus 5)
So d(Hi)=4(y-1)^3 and d(Ho)=-5(2y+1)(y^2+2y)^-6
and:
G'(y)=Hi x d(Ho) + Ho x d(Hi)
= (y-1)^4 x -5(2y+1)(y^2+2y)^-6 + (y^2+2y)^-5 x 4(y-1)^3

Answer 5

G (y)= (y-1)^4 / (y^2 + 2y) ^5
let u= (y-1)^4
thus, u'=4(y-1)^3

let v=(y^2+2y)^5
thus, v'= (10y+10)(y^2+2y)^4

G'(y)=(vu'-uv')/v^2( quotient rule)

to make it less complicated, use substitution method.
let y^2+2y=m and let y-1=n.

Thus,
G'(y)={(m^5)(4n^3)-[(n^4)(10n)(m^4)]} / (m^5)^2
=(4m^5n^3-10n^5m^4) / m^10
=[(4m^5n^3)/m^10] - [(10n^5m^4)/m^10]
=[(4n^3)/m^5]-[(10n^5)/m^6]
=(4mn^3-10n^5)/m^6
Now substitute y^2+2y=m and y-1=n into the equation.
so, G'(y)=[4(y^2+2y)(y-1)^3-10(y-i)^5] / (y^2+2y)^6

Answer 6

see... i'm somewhere below your level... let me guess...

y=y^4 +1/y^7 +2y^5... and solve from here... if i had equation editor i could tell you...

Answer 7

U GOT ME THERE.