Question

1. if ( f(x)=ln (x e^{-x}+e^{-x}) ), find ( f^{prime}(x) ).

a. ( \frac{-x}{1+x} )
b. ( \frac{1}{1-x} )
c. ( \frac{2-x}{1+x} )

Explanation:

Step1: Simplify the function

First, use the logarithmic property \(\ln(ab)=\ln a+\ln b\).
So \(f(x)=\ln(xe^{-x}+e^{-x})=\ln(e^{-x}(x + 1))\).
Since \(\ln(ab)=\ln a+\ln b\), then \(f(x)=\ln(e^{-x})+\ln(x + 1)\).
And \(\ln(e^{-x})=-x\), so \(f(x)=-x+\ln(x + 1)\).

Step2: Differentiate the function

Differentiate term - by - term.
The derivative of \(-x\) is \(-1\), and the derivative of \(\ln(u)\) with \(u=x + 1\) is \(\frac{u'}{u}\).
Since \(u=x + 1\), \(u'=1\).
By the sum rule of differentiation \((u + v)'=u'+v'\), \(f'(x)=-1+\frac{1}{x + 1}\).

Step3: Simplify the derivative

\(f'(x)=\frac{- (x + 1)+1}{x + 1}=\frac{-x-1 + 1}{x + 1}=\frac{-x}{x + 1}\).

Answer:

A. \(\frac{-x}{1 + x}\)