Find each derivative 3) \( y=\left(\frac{x}{3 x^{2}+2 x+1}\right)^{\frac{1}{3}} \)

We wish to differentiate   y = [x/(3x² + 2x + 1)]^(1/3). Step 1. Write y in terms of an inner function. Let   f(x) = x/(3x² + 2x + 1),   so that y = [f(x)]^(1/3). Step 2. Differentiate using the chain rule. The derivative of y with respect to x is   dy/dx = (1/3) [f(x)]^(–2/3) · f '(x). Step 3. Compute f '(x) using the quotient rule. Recall the quotient rule: if f(x) = u(x)/v(x), then   f '(x) = (u'v – uv') / [v(x)]². For our function,   u(x) = x → u' = 1,   v(x) = 3x² + 2x + 1 → v' = 6x + 2. Thus,   f '(x) = { (1)(3x² + 2x + 1) – x(6x + 2) } / (3x² + 2x + 1)². Simplify the numerator:   3x² + 2x + 1 – 6x² – 2x = –3x² + 1. So,   f '(x) = (–3x² + 1) / (3x² + 2x + 1)². Step 4. Combine the results. Substitute f(x) and f '(x) back into the chain rule formula:   dy/dx = (1/3) · [x/(3x² + 2x + 1)]^(–2/3) · [ (–3x² + 1) / (3x² + 2x + 1)² ]. Notice that the factor [x/(3x²+2x+1)]^(–2/3) can be rewritten as   [x/(3x²+2x+1)]^(–2/3) = [ (3x²+2x+1)/x ]^(2/3). Thus we have:   dy/dx = (1/3) · [ (3x²+2x+1)/x ]^(2/3) · [ (–3x² + 1) / (3x²+2x+1)² ]. Step 5. Simplify the expression. Combine the factors involving (3x² + 2x + 1):   [ (3x²+2x+1)^(2/3) ] / (3x²+2x+1)²     = 1 / (3x²+2x+1)^(4/3). Also, the x in the denominator contributes a factor of x^(2/3) from [ (3x²+2x+1)/x ]^(2/3). Thus, the derivative becomes:   dy/dx = (1/3) · (–3x² + 1) / [ x^(2/3)(3x²+2x+1)^(4/3) ]. It is often nice to write the numerator as (1 – 3x²). So the final derivative is:   dy/dx = (1 – 3x²) / [ 3·x^(2/3)·(3x²+2x+1)^(4/3) ]. This is the desired derivative.