Question

Differentiate.

\(f\left( t \right) = \frac{{\sqrt[3]{t}}}{{t - 3}}\)

Short answer

The answer is \(f'\left( t \right) = \frac{{ - 2t - 3}}{{3{t^{\frac{2}{3}}}{{\left( {t - 3} \right)}^2}}}\)

Step-by-step solution

Derivative of multiplication of two function

If a function isdivision of two function the we can differentiate the function by the following rule:

Let the function be \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) then the derivativewill be

\(\begin{aligned}h'\left( x \right) &= \frac{d}{{dx}}h\left( x \right)\\ &= \frac{d}{{dx}}\frac{{f\left( x \right)}}{{g\left( x \right)}}\\ &= \frac{{g\left( x \right)\frac{d}{{dx}} - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

Finding the derivative of the given function

The function is given by \(f\left( t \right) = \frac{{\sqrt[3]{t}}}{{t - 3}}\).

Differentiate the function with respect to \(t\) we get

\(\begin{aligned}f'\left( t \right) &= \frac{d}{{dt}}\left( {\frac{{\sqrt[3]{t}}}{{t - 3}}} \right)\\ &= \frac{{\left( {t - 3} \right)\frac{d}{{dt}}\left( {\sqrt[3]{t}} \right) - \sqrt[3]{t}\frac{d}{{dt}}\left( {t - 3} \right)}}{{{{\left( {t - 3} \right)}^2}}}\\ &= \frac{{\left( {t - 3} \right)\left( {\frac{1}{3} \cdot \frac{1}{{{t^{\frac{2}{3}}}}}} \right) - \sqrt[3]{t}}}{{{{\left( {t - 3} \right)}^2}}}\\ &= \frac{{\frac{1}{{3{t^{\frac{2}{3}}}}}\left( {t - 3 - 3{t^{\frac{1}{3}}}*{t^{\frac{2}{3}}}} \right)}}{{{{\left( {t - 3} \right)}^2}}}\\ &= \frac{{\left( { - 2t - 3} \right)}}{{3{t^{\frac{2}{3}}}{{\left( {t - 3} \right)}^2}}}\end{aligned}\)

Hence \(f'\left( t \right) = \frac{{ - 2t - 3}}{{3{t^{\frac{2}{3}}}{{\left( {t - 3} \right)}^2}}}\).