find the derivative of the function ( b(t)=sqrt3{\frac{t}{4 t^{2}+3}} ).\nanswer: ( b^{prime}(t)= )

Explanation:

Step1: Rewrite the function

Rewrite ( B(t)=\sqrt[3]{\frac{t}{4t^{2}+3}}=\left(\frac{t}{4t^{2}+3}\right)^{\frac{1}{3}}) using the chain - rule (y = u^{\frac{1}{3}}) where (u=\frac{t}{4t^{2}+3}). The chain - rule states that (y^\prime=\frac{1}{3}u^{-\frac{2}{3}}\cdot u^\prime).

Step2: Find the derivative of (u)

Use the quotient rule. If (u=\frac{f(t)}{g(t)}) where (f(t)=t) and (g(t)=4t^{2}+3), the quotient rule (u^\prime=\frac{f^\prime(t)g(t)-f(t)g^\prime(t)}{g(t)^{2}}). Since (f^\prime(t) = 1) and (g^\prime(t)=8t), then (u^\prime=\frac{(1)(4t^{2}+3)-t(8t)}{(4t^{2}+3)^{2}}=\frac{4t^{2}+3 - 8t^{2}}{(4t^{2}+3)^{2}}=\frac{3 - 4t^{2}}{(4t^{2}+3)^{2}}).

Step3: Combine using the chain - rule

(B^\prime(t)=\frac{1}{3}\left(\frac{t}{4t^{2}+3}\right)^{-\frac{2}{3}}\cdot\frac{3 - 4t^{2}}{(4t^{2}+3)^{2}}). Simplify (\left(\frac{t}{4t^{2}+3}\right)^{-\frac{2}{3}}=\left(\frac{4t^{2}+3}{t}\right)^{\frac{2}{3}}). So (B^\prime(t)=\frac{1}{3}\cdot\frac{4t^{2}+3}{t^{\frac{2}{3}}}\cdot\frac{3 - 4t^{2}}{(4t^{2}+3)^{2}}). Further simplify to (B^\prime(t)=\frac{3 - 4t^{2}}{3t^{\frac{2}{3}}(4t^{2}+3)^{\frac{4}{3}}}).

Answer:

(B^\prime(t)=\frac{3 - 4t^{2}}{3t^{\frac{2}{3}}(4t^{2}+3)^{\frac{4}{3}}})