Calculus hw help...DIFFERENTIATE problem!?

Question

Hey calculus geniuses...I need some assistance on this problem. Please solve it and factor it out completely THANKS!!
1) Differenciate. Answer must be completely factored.
y=(x+1)squared(x squared+1)cubed

THANKS!!!!!!!!

Answer 1

1) y = (x+1)^2(x^2+1)^3

Product rule:
y' = (x^2+1)^3*d/dx[(x+1)^2] + (x+1)^2*d/dx[(x^2+1)^3]
= (x^2+1)^3(2)(x+1) + (x+1)^2(3)(x^2+1)^2(2x)
= 2(x^2+1)^3(x+1) + 6x(x+1)^2(x^2+1)^2
= (x^2+1)^2(x+1)[2(x^2+1) + 6x(x+1)]
= (x^2+1)^2(x+1)[2x^2+2 + 6x^2+6x]
= (x^2+1)^2(x+1)[8x^2 + 6x + 2]
= 2(x^2+1)^2(x+1)[4x^2 + 3x + 1]

The last part, 4x^2 + 3x + 1, cannot be factored, unless you allow imaginary numbers.

Answer 2

do no longer put in 9 in the given equation as others have pronounced. P(x) is the equation that supplies the inhabitants. the 1st by-made from any equation is generally defined as "the value of exchange of the equation". So, you want to take the 1st by-made from P(x) to get P'(x) (in simple terms yet another notation for the 1st by-made from P(x)). this could nicely be an elementary derivative wherein the hassle-loose capability rule applies. After differentiation, then you definately can positioned 9 in for x in P'(x) and you gets the value of exchange of the inhabitants after 9 months.

Answer 3

y=(x+1)^2(x^2+1)^3
y'=(x+1)^2*6x(x^2+1)^2
+(x^2+1)^3(2(x+1)
=6x(x+1)^2(x^2+1)^2
+2(x+1)(x^2+1)^3
=2(x+1)(x2+1)^2[3x(x+1)+x^2+1)]
=2(x+1)(x^2+1)^2[4x^2+3x+1]