Final answer:
To find g'(1) for the function g(x) = 3x²/(3x + 1), you would use the quotient rule to differentiate and then substitute x=1 into the result. If the +1 is not in the denominator, then g'(1) is simply 2.
Step-by-step explanation:
To find the derivative of the function g(x) = ³x²/3x + 1 at x = 1 using the quotient rule, we must first correct the notation to g(x) = ³x²/(3x + 1) assuming the "+1" is in the denominator; if not, the problem simplifies to g'(x) = 6x/3 = 2x and g'(1) = 2.
If the assumption is correct and "+1" is part of the denominator, the quotient rule is as follows: ²g²(x) = [v(x)²u²(x) - u(x)²v²(x)] / [v(x)]² where u(x) is the numerator and v(x) is the denominator of the function.
The derivative ²g²(x) is then: ²g²(x) = [(3x + 1)(2x) - x²(3)] / (3x + 1)² = (3x² + 6x - 3x²) / (3x + 1)² = 6x / (3x + 1)². Therefore, g'(1) = 6 / (3 * 1 + 1)² = 6 / 16 = 3 / 8.