f (x) = (3x2-x)4 (x3+4x)3 I don't know if this is written right.
W wjunior30 New member Joined Nov 28, 2006 Messages 1 Nov 28, 2006 #1 f (x) = (3x2-x)4 (x3+4x)3 I don't know if this is written right.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Nov 28, 2006 #2 neither do I ... do you mean f(x)=(3x2−x)4(x3+4x)3\displaystyle f(x) = (3x^2-x)^4 (x^3+4x)^3f(x)=(3x2−x)4(x3+4x)3 ??? if so, the product and chain rules are in order ... f′(x)=(3x2−x)4⋅3(x3+4x)2⋅(3x2+4)+(x3+4x)3⋅4(3x2−x)3⋅(6x−1)\displaystyle f'(x) = (3x^2-x)^4 \cdot 3(x^3+4x)^2 \cdot (3x^2 + 4) + (x^3+4x)^3 \cdot 4(3x^2-x)^3 \cdot (6x-1)f′(x)=(3x2−x)4⋅3(x3+4x)2⋅(3x2+4)+(x3+4x)3⋅4(3x2−x)3⋅(6x−1) to simplify, factor out common factors from both terms ... f′(x)=(3x2−x)3(x3+4x)2[3(3x2−x)(3x2+4)+4(x3+4x)(6x−1)]\displaystyle f'(x) = (3x^2-x)^3(x^3+4x)^2[3(3x^2-x)(3x^2+4) + 4(x^3+4x)(6x-1)]f′(x)=(3x2−x)3(x3+4x)2[3(3x2−x)(3x2+4)+4(x3+4x)(6x−1)] I'll leave the remaining algebra "clean-up" for you.
neither do I ... do you mean f(x)=(3x2−x)4(x3+4x)3\displaystyle f(x) = (3x^2-x)^4 (x^3+4x)^3f(x)=(3x2−x)4(x3+4x)3 ??? if so, the product and chain rules are in order ... f′(x)=(3x2−x)4⋅3(x3+4x)2⋅(3x2+4)+(x3+4x)3⋅4(3x2−x)3⋅(6x−1)\displaystyle f'(x) = (3x^2-x)^4 \cdot 3(x^3+4x)^2 \cdot (3x^2 + 4) + (x^3+4x)^3 \cdot 4(3x^2-x)^3 \cdot (6x-1)f′(x)=(3x2−x)4⋅3(x3+4x)2⋅(3x2+4)+(x3+4x)3⋅4(3x2−x)3⋅(6x−1) to simplify, factor out common factors from both terms ... f′(x)=(3x2−x)3(x3+4x)2[3(3x2−x)(3x2+4)+4(x3+4x)(6x−1)]\displaystyle f'(x) = (3x^2-x)^3(x^3+4x)^2[3(3x^2-x)(3x^2+4) + 4(x^3+4x)(6x-1)]f′(x)=(3x2−x)3(x3+4x)2[3(3x2−x)(3x2+4)+4(x3+4x)(6x−1)] I'll leave the remaining algebra "clean-up" for you.
