Question

Differentiate the function.

21.\(y = \ln \left( {{e^{ - x}} + x{e^{ - x}}} \right)\)

Short answer

The derivative of the function is \(y' = - \frac{x}{{1 + x}}\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Differentiate the function

Rewrite the function as shown below:

\(\begin{aligned}{c}y&= \ln \left( {{e^{ - x}} + x{e^{ - x}}} \right)\\&= \ln \left( {{e^{ - x}}\left( {1 + x} \right)} \right)\\&= \ln \left( {{e^{ - x}}} \right) + \ln \left( {1 + x} \right)\\&= - x + \ln \left( {1 + x} \right)\end{aligned}\)

Use the above formula to differentiate the above function as shown below:

\(\begin{aligned}{c}y'&= \frac{d}{{dx}}\left( { - x + \ln \left( {1 + x} \right)} \right)\\&= \frac{d}{{dx}}\left( { - x} \right) + \frac{d}{{dx}}\ln \left( {1 + x} \right)\\&= - 1 + \frac{1}{{1 + x}} \cdot \left( {0 + 1} \right)\\&= \frac{{ - 1 - x + 1}}{{1 + x}}\\&= - \frac{x}{{1 + x}}\end{aligned}\)

Thus, the derivative of the function is \(y' = - \frac{x}{{1 + x}}\).