Find the derivative of the function \( B(t)=\sqrt[3]{\frac{t}{3 t^{2}+4}} \)

We are given the function   B(t) = ∛(t/(3t² + 4)) = [t/(3t² + 4)]^(1/3). To find its derivative B′(t), we will use the chain rule along with the quotient rule. Follow these steps: Step 1. Write B(t) as a power function:   B(t) = [u(t)]^(1/3),  where u(t) = t/(3t² + 4). By the chain rule,   B′(t) = (1/3) [u(t)]^(1/3 − 1) · u′(t) = (1/3) [u(t)]^(–2/3) · u′(t). Step 2. Compute u′(t) using the quotient rule. Recall that if u(t) = f(t)/g(t), then   u′(t) = [g(t) f′(t) − f(t) g′(t)] / [g(t)]². Here, f(t) = t and g(t) = 3t² + 4, so   f′(t) = 1  and  g′(t) = 6t. Thus,   u′(t) = [(3t² + 4)(1) − t(6t)] / (3t² + 4)²        = [3t² + 4 − 6t²] / (3t² + 4)²        = (4 − 3t²) / (3t² + 4)². Step 3. Substitute u(t) and u′(t) back into the expression for B′(t):   B′(t) = (1/3) [t/(3t² + 4)]^(–2/3) · [(4 − 3t²)/(3t² + 4)²]. To simplify the expression [t/(3t² + 4)]^(–2/3), note that a negative exponent inverts the fraction:   [t/(3t² + 4)]^(–2/3) = (3t² + 4)^(2/3)/t^(2/3). Now substitute this back:   B′(t) = (1/3) · [(3t² + 4)^(2/3)/t^(2/3)] · [(4 − 3t²)/(3t² + 4)²]. Combine the factors involving (3t² + 4):   (3t² + 4)^(2/3) / (3t² + 4)² = 1/(3t² + 4)^(2 − 2/3) = 1/(3t² + 4)^(4/3). Thus, the derivative simplifies to:   B′(t) = (1/3) · (4 − 3t²) / [t^(2/3) (3t² + 4)^(4/3)]. So the final answer is:   B′(t) = (4 − 3t²) / [3 · t^(2/3) · (3t² + 4)^(4/3)].