find F(x)' for x^3(3-x)^4.. i think answer is 3x^2(3-x)(3-3x)^4.. please help thanks

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

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)

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.

To find the derivative of the given function, we can make use of the product rule. The product rule states that if we have two functions, f(x) and g(x), then the derivative of their product is given by:

(f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)

Now let's find the derivatives of the individual functions:

f(x) = x^3
Using the power rule, the derivative of x^n (where n is a constant) is given by n*x^(n-1). Applying this rule, we find:
f'(x) = 3x^2

g(x) = (3-x)^4
To differentiate this function, we can use the chain rule. The chain rule states that if we have a composite function, y = f(g(x)), then the derivative is given by:

dy/dx = f'(g(x)) * g'(x)

Let's differentiate g(x) step by step:
g'(x) = -1 * 4(3-x)^3
= -4(3-x)^3

Now we can find the derivative of the original function using the product rule:

F(x) = f(x) * g(x) = x^3 * (3-x)^4

F'(x) = f'(x) * g(x) + f(x) * g'(x)
= (3x^2) * (3-x)^4 + (x^3) * (-4)(3-x)^3

Simplifying the expression further, we have:
F'(x) = 3x^2(3-x)^4 - 4x^3(3-x)^3

So, the correct derivative of F(x) is given by:
F'(x) = 3x^2(3-x)^4 - 4x^3(3-x)^3