Find the derivatives: A).Y=(X-1)3 (x+2)4 B).Y=(X2+ 1)5 C).Y=X3-3X)4 D.) Y=(X+1)2 (X2+1) -3?

Product rule: (f*g)'= f*g' + g*f'

Chain rule: ((f)^n)' = n*f^(n-1)*f'

y=(x-1)^3 * (x+2)^4

y'=(x-1)^3 * 4(x+2)^3 dx + (x+2)^4 * 3(x-1)^2 dx

y'=(x-1)^2 * (x+2)^3 *[4(x-1) + 3(x+2)]dx

y'=(x-1)^2 * (x+2)^3 * (7x+2) dx

y=(x^2+1)^5

y'= 5(x^2+1)^4 * 2xdx

y'=10x(x^2+1)^4 dx

y=x^3 - 3x

y'= (3x^2 -3) dx

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

y'=(x+1)^2 * (-3)2x(x^2+1)^-4 dx + (x^2 +1)^-3 * 2(x+1)dx

y'=2(x+1)(x^2+1)^-4 * [-3x(x+1) + x^2+1]dx

y'= 2(x+1)(x^2+1)^-4 * (-2x^2 -3x +1)dx

A: 4(x-1)^3 (x+2)^3 + 3(x+2)^4 (x-1)^2 , which simplifies to:

4x^6 + 12x^5 - 12x^4 - 44x^3 +24x^2 + 48x - 32

B: 5 (x^2 + 1)^4 (2x) which equals 10x (x^2 +1)^4

C: 4 ((x^3 - 3x)^3) (3x^2 - 3) which equals

12 (x^2-1) ((x^3 - 3x)^3)

D: ( (2x+2) / ((x^2 + 1)^3) ) - ( (6x^3 - 12x^2 - 6x) / ((x^2 + 1)^4) )

wow that was ugly. hope this helps.

dam that reminds me when i was in AP calculus use to do those so easy but i dont quite remember how to do them anymore :( cant beleave at one point those were so ez to me

dunno