product rule.
F(x)= f(x)*g(x)
F'(x)= f'(x)g(x)+f(x)g'(x)
so your answer would be 3x^2(3-x)^4-4x^3(3-x)^2
find F(x)' for x^3(3-x)^4.. i think answer is 3x^2(3-x)(3-3x)^4.. please help thanks
4 answers
no
using the product rule, I got
dy/dx = x^3(4)(3-x)^3 (-1) + (3-x)^4 (3x^2)
= x^2 (3-x)^3 [-4x + 3(3-x) ]
= x^2 (3-x)^3 (9-7x)
using the product rule, I got
dy/dx = x^3(4)(3-x)^3 (-1) + (3-x)^4 (3x^2)
= x^2 (3-x)^3 [-4x + 3(3-x) ]
= x^2 (3-x)^3 (9-7x)
they're the same answer. you just simplified it.
oh right it should be 3x^2(3-x)^4-4x^3(3-x)^3, mistyped that.