Question

Differentiate.

\(W\left( t \right) = {e^t}\left( {1 + t{e^t}} \right)\)

Short answer

The answer is \(W'\left( t \right) = {e^t}\left( {1 + {e^t} + 2t{e^t}} \right)\)

Step-by-step solution

Derivative of multiplication of two function

If a function is multiplicationof two function then 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 derivative will 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 given function is \(W\left( t \right) = {e^t}\left( {1 + t{e^t}} \right)\). Differentiating the function we get

\(\begin{aligned}W'\left( t \right) &= \frac{d}{{dt}}\left( {{e^t}\left( {1 + t{e^t}} \right)} \right)\\ &= {e^t}\frac{d}{{dt}}\left( {1 + t{e^t}} \right) + \left( {1 + t{e^t}} \right)\frac{d}{{dt}}{e^t}\\ &= {e^t}\left( {{e^t} + t{e^t}} \right) + \left( {1 + t{e^t}} \right){e^t}\\ &= {e^t}\left( {1 + {e^t} + 2t{e^t}} \right)\end{aligned}\)

Hence \(W'\left( t \right) = {e^t}\left( {1 + {e^t} + 2t{e^t}} \right)\).