Question

Differentiate the function.

9.\(g\left( x \right) = \ln \left( {x{e^{ - 2x}}} \right)\).

Short answer

The differentiation of\(g\left( x \right) = \ln \left( {x{e^{ - 2x}}} \right)\) is \(\frac{1}{x} - 2\).

Step-by-step solution

Formula used

1. Logarithmic formula: \(\ln \left( {ab} \right) = \ln \left( a \right) + \ln \left( b \right)\)
2. Logarithmic property: \(\ln {e^n} = n\)
3. Differential formula:\(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\)

Simplify the given function

Simplify the given function \(g\left( x \right) = \ln \left( {x{e^{ - 2x}}} \right)\) by using logarithm formula \(\ln \left( {ab} \right) = \ln \left( a \right) + \ln \left( b \right)\).

\(g\left( x \right) = \ln x + \ln \left( {{e^{ - 2x}}} \right)\)

Use \(\ln {e^n} = n\) in the obtained function.

\(g\left( x \right) = \ln x - 2x\)

Differentiate the given function

Differentiate\(g\left( x \right) = \ln x - 2x\) with respect to \(x\) by using\(\frac{d}{{dx}}\left( {\ln u} \right)= \frac{1}{u}\frac{{du}}{{dx}}\).

\(\begin{aligned}{c}g'\left( x \right)&= \frac{d}{{dx}}\left( {\ln x - 2x} \right)\\&= \frac{d}{{dx}}\left( {\ln x} \right) - \frac{d}{{dx}}\left( {2x} \right)\\&= \frac{1}{x} - 2\end{aligned}\)

Hence, the differentiation of \(g\left( x \right) = \ln \left( {x{e^{ - 2x}}} \right)\) is \(\frac{1}{x} - 2\).