Question

Differentiate.

\(F\left( z \right) = \left( {1 - {e^z}} \right)\left( {z + {e^z}} \right)\)

\(\)

Short answer

Derivative of the function is \(\left( {1 - z{e^z} - 2{e^{2z}}} \right)\)\(\).

Step-by-step solution

Precise Definition of differentiation

A derivative is a function of a real variable.It is the rate of change of output value with respect to an input value. The derivative function is a single variable at a selected input value.

Find the derivative of F using the Product rule

The equation for the Product rule is \({\left( {fg} \right)^\prime } = f'g + g'f\).

Apply this rule as:

\(\begin{aligned}\frac{d}{{dz}}F\left( z \right) &= \frac{d}{{dz}}\left( {\left( {1 - {e^z}} \right)\left( {z + {e^z}} \right)} \right)\\ &= \left( {1 - {e^z}} \right)\frac{d}{{dz}}\left( {z + {e^z}} \right) + \left( {z + {e^z}} \right)\frac{d}{{dz}}\left( {1 - {e^z}} \right)\end{aligned}\)

Simplify the obtained condition

Apply derivative rule in obtained condition as:

\(\begin{aligned}F'\left( z \right) &= \left( {1 - {e^z}} \right)\frac{d}{{dz}}\left( {z + {e^z}} \right) + \left( {z + {e^z}} \right)\frac{d}{{dz}}\left( {1 - {e^z}} \right)\\ &= \left( {1 - {e^z}} \right)\left( {1 + {e^z}} \right) + \left( {z + {e^z}} \right)\left( { - {e^z}} \right)\\ &= \left( {1 - {e^z} + {e^z} - {e^{2z}}} \right) + \left( { - z{e^z} - {e^{2z}}} \right)\\ &= \left( {1 - z{e^z} - 2{e^{2z}}} \right)\end{aligned}\)

So, \(F'(z) = \left( {1 - z{e^z} - 2{e^{2z}}} \right)\)is the final answer of the derivative function.