quick help maths (1 Viewer)

[shady145]

find the second derivative
x/(x+1)
i got x(x+1)^-1
y'=-x(x+1)^-2
y''=2x(x+1)^-3
y''=2x/(x+1)^3
but the answer is -2x/(x+1)^3

Now is this the right method coz i initially thought that i had to use the product rule doing my second derivative because there was an x term out the front.

Can some1 please tell me the right way
thankyou

 

[kaz1]

Quotient rule.
y'=[(x+1)-x]/(x+1)²
y'=1/(x+1)²
y'=(x+1)^-2
Chain rule.
y''=-2(x+1)^-3
y''=-2/(x+1)³

but the answer is -2x/(x+1)^3

Fuck. Can someone point out my mistake?

 

[shady145]

The product rule...

u=x v=(x+1)^-1
u'=1 v'=-(x+1)^-2

y'=(x+1)^-1*(1) + (x)*-(x+1)^-2
y'=(x+1)^-1 -x(x+1)^-2

taking -x(x+1)^-2
u=-x v=(x+1)^-2
u'=-1 v'=-2(x+1)^-3

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

 

[shady145]

ooo thank you
kaz1

 

[shady145]

haha dont worry i know i did the product rule wrong

 

[youngminii]

Quotient rule.
y'=[(x+1)-x]/(x+1)²
y'=1/(x+1)²
y'=(x+1)^-2
Chain rule.
y''=-2(x+1)^-3
y''=-2/(x+1)³



Fuck. Can someone point out my mistake?

No mistake, you're right.

 

[Trebla]

You can actually avoid the quotient/product rule (which is very prone to silly mistakes) if you note that:
x/(x + 1) = (x + 1 - 1)/(x + 1) = 1 - 1/(x + 1)
which becomes a simpler form to differentiate without any product or quotient rule...lol

 

[youngminii]

You can actually avoid the quotient/product rule (which is very prone to silly mistakes) if you note that:
x/(x + 1) = (x + 1 - 1)/(x + 1) = 1 - 1/(x + 1)
which becomes a simpler form to differentiate without any product or quotient rule...lol

Genius.