Question

Find the derivative of the function $f\left(x\right)=x^x$

• A. $f\left(x\right)\cdot [1–\ln x]$
• B. $f\left(x\right)\cdot [1+\ln x]$
• C. $f\left(x\right)\cdot \ln x$
• D. $f\left(x\right)-\ln x$

Answer 1

The correct option is B $f\left(x\right)\cdot [1+\ln x]$

Given function is,

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

This is a complex looking function compared to our standard functions. Logarithmic differentiation can be used to tackle such situation.

It simply means taking the logarithm before we actually differentiate. Take logarithm to simplify the function if it is of the form,

$f\left(x\right)=g(x{)}^{h\left(x\right)}$

when we take logarithm, the power becomes a multiplier as below,

$\log \left[f\right(x\left)\right]=h\left(x\right)\cdot \log \left[g\right(x\left)\right]$

Differentiating is easier now. Logarithmic differentiation can be used in other scenarios as well especially when the function is a quotient of 2 other functions.

In our present ease,

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

Taking log on both sides

$lnf\left(x\right)=x\cdot \ln x$

Differentiating both sides

$\frac{1}{f\left(x\right)}\cdot f^{\prime }\left(x\right)=x\cdot \frac{1}{x}+\ln x$
(note that we are using chain rule to differentiate the LHS)

$\frac{1}{f\left(x\right)}\cdot f^{\prime }\left(x\right)=1+\ln x\Rightarrow f^{\prime }\left(x\right)=x^x(1+\ln x)$

So option B is the correct answer.