Question

Differentiate.

\(V\left( t \right) = \left( {t + 2{e^t}} \right)\sqrt t \)

Short answer

The answer is \(V'\left( t \right) = \frac{{3t + 2{e^t} + 4t{e^t}}}{{2\sqrt t }}\).

Step-by-step solution

Derivative of multiplication of two function

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

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

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

Finding the derivative of the given function

The function is given by \(V\left( t \right) = \left( {t + 2{e^t}} \right)\sqrt t \).

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

\(\begin{aligned}V'\left( t \right) &= \frac{d}{{dt}}\left( {\left( {t + 2{e^t}} \right)\sqrt t } \right)\\ &= \left( {t + 2{e^t}} \right)\frac{d}{{dt}}\sqrt t + \sqrt t \frac{d}{{dt}}\left( {t + 2{e^t}} \right)\\ &= \left( {t + 2{e^t}} \right)*\frac{1}{2}\frac{1}{{\sqrt t }} + \sqrt t \left( {1 + 2{e^t}} \right)\\ &= \frac{{\left( {t + 2{e^t}} \right)}}{{2\sqrt t }} + \sqrt t \left( {1 + 2{e^t}} \right)\\ &= \frac{{\left( {t + 2{e^t}} \right) + 2t\left( {1 + 2{e^t}} \right)}}{{2\sqrt t }}\\ &= \frac{{3t + 2{e^t} + 4t{e^t}}}{{2\sqrt t }}\end{aligned}\)

Hence \(V'\left( t \right) = \frac{{3t + 2{e^t} + 4t{e^t}}}{{2\sqrt t }}\).