differentiate.\n$f(t)=\frac{sqrt3{t}}{t - 3}$\n$f(t)=$

Explanation:

Step1: Rewrite the function

Rewrite $\sqrt[3]{t}$ as $t^{\frac{1}{3}}$, so $f(t)=\frac{t^{\frac{1}{3}}}{t - 3}$. Use the quotient - rule which states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = t^{\frac{1}{3}}$ and $v=t - 3$.

Step2: Find $u'$ and $v'$

Differentiate $u=t^{\frac{1}{3}}$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$. So $u'=\frac{1}{3}t^{\frac{1}{3}-1}=\frac{1}{3}t^{-\frac{2}{3}}$. And $v'=\frac{d}{dt}(t - 3)=1$.

Step3: Apply the quotient - rule

$f'(t)=\frac{u'v - uv'}{v^{2}}=\frac{\frac{1}{3}t^{-\frac{2}{3}}(t - 3)-t^{\frac{1}{3}}\times1}{(t - 3)^{2}}$.

Step4: Simplify the numerator

Expand the numerator: $\frac{1}{3}t^{-\frac{2}{3}}(t - 3)-t^{\frac{1}{3}}=\frac{1}{3}t^{-\frac{2}{3}}\times t-\frac{1}{3}t^{-\frac{2}{3}}\times3 - t^{\frac{1}{3}}$. Using the rule $a^{m}\times a^{n}=a^{m + n}$, we have $\frac{1}{3}t^{-\frac{2}{3}+1}-t^{-\frac{2}{3}}-t^{\frac{1}{3}}=\frac{1}{3}t^{\frac{1}{3}}-t^{-\frac{2}{3}}-t^{\frac{1}{3}}$. Combine like - terms: $\frac{1}{3}t^{\frac{1}{3}}-t^{\frac{1}{3}}-t^{-\frac{2}{3}}=-\frac{2}{3}t^{\frac{1}{3}}-t^{-\frac{2}{3}}$. So $f'(t)=\frac{-\frac{2}{3}t^{\frac{1}{3}}-t^{-\frac{2}{3}}}{(t - 3)^{2}}=\frac{-\frac{2}{3}\sqrt[3]{t}-\frac{1}{\sqrt[3]{t^{2}}}}{(t - 3)^{2}}=\frac{-\frac{2t - 3}{3\sqrt[3]{t^{2}}}}{(t - 3)^{2}}=\frac{3 - 2t}{3\sqrt[3]{t^{2}}(t - 3)^{2}}$.

Answer:

$\frac{3 - 2t}{3\sqrt[3]{t^{2}}(t - 3)^{2}}$