The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

differentiation

given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

dy/dx= ax^2 + 2bx / (ax+b)^2

where do i go from here
Original post by Custardcream000
given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

dy/dx= ax^2 + 2bx / (ax+b)^2

where do i go from here


Assuming it's correct (I can't be bothered to check) then just differentiate it again.
Original post by RDKGames
Assuming it's correct (I can't be bothered to check) then just differentiate it again.


so d^2y/dx^2 = [(ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))]/ (ax+b)^4
(edited 7 years ago)
Original post by Custardcream000
so d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a)/ (ax+b)^4


Nope. Check it again and post your working if you get stuck.
Original post by RDKGames
Nope. Check it again and post your working if you get stuck.


d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
Original post by Custardcream000
d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4


No...

You shouldn't even get a power of 4.
Original post by Custardcream000
d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4


It might be easier to follow if you work it out on paper and then post a picture of your working.
Original post by RDKGames
No...

You shouldn't even get a power of 4.


i'm using the (vu' - uv')/v^2 method though
(edited 7 years ago)
please put brackets around the top of the fraction.
Original post by Custardcream000
i'm using the vu' - uv'/v^2 method though


In which case one power should cancel and it's hard to see where without your working.

u=ax2+2bx⇒u′=2ax+2bu=ax^2+2bx \Rightarrow u'=2ax+2b

v=(ax+b)2⇒v′=2a(ax+b)v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

vu′−uv′v2=(2ax+2b)(ax+b)2−(ax2+bx)[2a(ax+b)](ax+b)4\displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

You can factor out one ax+bax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.
(edited 7 years ago)
Original post by RDKGames
In which case one power should cancel and it's hard to see where without your working.

u=ax2+2bx⇒u′=2ax+2bu=ax^2+2bx \Rightarrow u'=2ax+2b

v=(ax+b)2⇒v′=2a(ax+b)v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

vu′−uv′v2=(2ax+2b)(ax+b)2−(ax2+bx)[2a(ax+b)](ax+b)4\displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

You can factor out one ax+bax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.


thanks so i get

{2a^3x^2 + 4a^2bx^2 + 6ab^2x +2b^3 - 2a^2x^2 + 2abx} / (ax+b)^3

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you