Question

Why do we need to consider absolute values when we apply logarithmic differentiation to $f\left(x\right)=x{e}^x\sin x?$In contrast, why do we not need to consider absolute values when we apply logarithmic differentiation to $f\left(x\right)=x^x?$

Short answer

In $f\left(x\right)=x^x,$log turns power into product and product into sum but in localid="1649006792512" $f\left(x\right)=x{e}^x\sin x$there is no need to be logarithmic differentiation.

Step-by-step solution

Step 1. Introduction

We need to write why we need to consider absolute values when we apply logarithmic differentiation to$f\left(x\right)=x{e}^x\sin x$ and also why we need to consider absolute values when we apply logarithmic differentiation to $f\left(x\right)=x^x.$

Step 2. Explanation

In calculus logarithmic differentiation refers to the process of first taking the natural log of a function $y=f\left(x\right)$then solving for the derivative $\frac{dy}{dx}.$On a surface of it, it would seem that logs would only make a complicated function more complicated but that logs turn power into a product and product into sums, like,

$f\left(x\right)=x^x.$

Take log both sides.

$\log f\left(x\right)=\log x^x.$

$\log f\left(x\right)=x\log x.$

Differentiating both sides.

$\frac{1}{f\left(x\right)}\times f^{\text{'}}\left(x\right)=\log x+x\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right..$

$f^{\text{'}}\left(x\right)=\left(1+\log x\right)f\left(x\right).$

$=\left(1+\log x\right)x^x.$

In $f\left(x\right)=x{e}^x\sin x,$we simply apply product rule.

$f\left(x\right)=x{e}^x\sin x.$

$f^{\text{'}}\left(x\right)=e^x\sin x+x{e}^x\sin x+x{e}^x\cos x.$