2nd Derivative Problem

[calculus 1983]

a quick question ...

i am mid-way through a problem and i am just curious why you distribute the (-4x^2) only to the 2 in --> (2(x^2+1))

this is what the problem looks like as of now ...that was just the piece i took from it

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

and since you only distributed the (-4x^2) to the 2 in (2(x^2+1))
the next step looks like this ...

= -2(x^2 + 1)^2 - (-8x^2 (x^2+1)) / (x^2 + 1)^4

 

[skeeter]

a quick question ...

i am mid-way through a problem and i am just curious why you distribute the (-4x^2) only to the 2 in --> (2(x^2+1))

let a = -4x², b = 2, and c = x² + 1 ...

is a(bc) = (ab)c, or is it a(bc) = ab*ac ?

what property of multiplication is this (that you learned long ago ... and, maybe, far away) ???

 

[calculus 1983]

lol it's been awhile ...
so then you would do (ab)c

or (-4x^2)(2) = -8x^2(x^2 + 1)