# differentiation

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#1
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
1
5 years ago
#2
(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.
0
#3
(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
0
5 years ago
#4
(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.
0
#5
(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
0
5 years ago
#6
(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.
0
5 years ago
#7
(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.
0
#8
(Original post by RDKGames)
No...

You shouldn't even get a power of 4.
i'm using the (vu' - uv')/v^2 method though
0
5 years ago
#9
please put brackets around the top of the fraction.
0
5 years ago
#10
(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.   You can factor out one 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.
0
#11
(Original post by RDKGames)
In which case one power should cancel and it's hard to see where without your working.   You can factor out one 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
0
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