Question

Differentiate $x^x$ from first principle where $x>0.$

• A. $x^x[1+\log x]$
• B. $x^x$
• C. $x^{x-1}$
• D. none of these

Answer 1

The correct option is A $x^x[1+\log x]$

$\frac{dy}{dx}=\underset{h\to 0}{lim}\frac{{(x+h)}^{x+h}-x^x}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}[{(x+h)}^{x+h}-x^{x+h}+x^{x+h}-x^x]$

$=\underset{h\to 0}{lim}\frac{1}{h}[x^{x+h}\{{\left(\frac{x+h}{x}\right)}^x.{\left(\frac{x+h}{x}\right)}^h-1\}+x^x(x^h-1)]$

$=x^x\underset{h\to 0}{lim}\frac{1}{h}[{\left\{{(1+\frac{h}{x})}^{x/h}\right\}}^h-1]+Lt\left[\frac{x^h-1}{h}\right]$

$=x^x[lim\frac{e^b-1}{h}+Lt\frac{x^h-1}{h}]$

$=x^x[\log e+\log x]$ $=x^x[1+\log x]$

$\because \underset{h\to 0}{lim}\frac{a^x-1}{x}=\log a$