Question

Differentiate.

28. \(F\left( t \right) = \frac{{At}}{{B{t^2} + C{t^3}}}\)

Short answer

The answer is \(F'\left( t \right) = - \frac{{A\left( {B + 2Ct} \right)}}{{{t^2}{{\left( {B + Ct} \right)}^2}}}\)

Step-by-step solution

Derivative of multiplication of two function

If a function is division 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 \(F\left( t \right) = \frac{{At}}{{B{t^2} + C{t^3}}}\).

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

\(\begin{aligned}F'\left( t \right) & = \frac{d}{{dt}}\left( {\frac{{At}}{{B{t^2} + C{t^3}}}} \right)\\ & = \frac{{\left( {B{t^2} + C{t^3}} \right)\frac{d}{{dt}}\left( {At} \right) - At\frac{d}{{dt}}\left( {B{t^2} + C{t^3}} \right)}}{{{{\left( {B{t^2} + C{t^3}} \right)}^2}}}\\ & = \frac{{A\left( {B{t^2} + C{t^3}} \right) - At\left( {2Bt + 3C{t^2}} \right)}}{{{t^4}{{\left( {B + Ct} \right)}^2}}}\\ & = \frac{{A{t^2}\left( {B + Ct - 2B - 3Ct} \right)}}{{{t^4}{{\left( {B + Ct} \right)}^2}}}\\ & = \frac{{ - A\left( {B + 2Ct} \right)}}{{{t^2}{{\left( {B + Ct} \right)}^2}}}\end{aligned}\)

Hence \(F'\left( t \right) = - \frac{{A\left( {B + 2Ct} \right)}}{{{t^2}{{\left( {B + Ct} \right)}^2}}}\).