Question

differentiate x^x by first principle

Answer 1

y = x^x
= e^xlogx

now differentiate with respect to x
we Know ,
deferentiation of e^ax
e.g e^ax .a , use this concept here,

dy/dx = e^xlogx. d(xlogx )/dx
= x^x {x.d(logx)/dx + logx.dx/dx }
[ x^x = e^xlogx used above ]
= x^x{1 + logx }

hence, deffentiation of x^x =x^x (1 + logx)

Answer 2

SOLUTION

TO DETERMINE

To differentiate by first principle

$\sf {x}^{x}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf 1. \: \: \: f'(x) = \lim_{h \to 0} \: \frac{f(x + h) - f(x)}{h}$

$\displaystyle \sf 2. \: \: \: \lim_{x \to 0} \: \frac{ {a}^{x} - 1}{x} = \ln a$

EVALUATION

Let f(x) be the given function

Then we have

$\sf f(x) = {x}^{x}$

Now using first principle on derivatives we get

$\displaystyle \sf f'(x)$

$\displaystyle \sf = \lim_{h \to 0} \: \frac{f(x + h) - f(x)}{h}$

$\displaystyle \sf = \lim_{h \to 0} \: \frac{ {(x + h)}^{x + h} - {x}^{x} }{h}$

$\displaystyle \sf = \lim_{h \to 0} \: \frac{ {(x + h)}^{x + h} - {x }^{x + h} + {x }^{x + h}- {x}^{x} }{h}$

$\displaystyle \sf = \lim_{h \to 0} \: \frac{ {(x + h)}^{x + h} - {x }^{x + h} + {x }^{x + h}- {x}^{x} }{h}$

$\displaystyle \sf = \lim_{h \to 0} \: \frac{ {(x + h)}^{x + h} - {x }^{x + h}}{h} + \lim_{h \to 0} \: \frac{ {x }^{x + h}- {x}^{x} }{h}$

$\displaystyle \sf = \lim_{h \to 0} \: {x}^{x + h}. \frac{ { \bigg( \dfrac{x + h}{x} \bigg)}^{x + h} - 1}{h} + \lim_{h \to 0} \: {x}^{x} . \frac{ {x }^{x + h}- 1}{h}$

$\displaystyle \sf = \lim_{h \to 0} \: {x}^{x + h}. \lim_{h \to 0} \frac{ { \bigg( \dfrac{x + h}{x} \bigg)}^{x + h} - 1}{h} + \: {x}^{x} . \lim_{h \to 0} \frac{ {x }^{h}- 1}{h}$

$\displaystyle \sf = {x}^{x }. \lim_{h \to 0} \frac{{ \bigg( 1 + \dfrac{ h}{x} \bigg)}^{x} { \bigg( 1 + \dfrac{ h}{x} \bigg)}^{ h} - 1}{h} + \: {x}^{x} . \lim_{h \to 0} \frac{ {x }^{ h}- 1}{h}$

$\displaystyle \sf = {x}^{x }. \lim_{h \to 0} \frac{ { \bigg( 1 + \dfrac{ h}{x} \bigg)}^{ x} .1- 1}{h} + \: {x}^{x} . \lim_{h \to 0} \frac{ {x }^{ h}- 1}{h}$

$\displaystyle \sf = {x}^{x }. \lim_{h \to 0} \frac{ \bigg[{{ \bigg( 1 + \dfrac{ h}{x} \bigg)}^{ \frac{x}{h} } \bigg] }^{h} - 1}{h} + \: {x}^{x} . \lim_{h \to 0} \frac{ {x }^{ h}- 1}{h}$

$\displaystyle \sf = {x}^{x }. \lim_{h \to 0} \frac{ {e}^{h} - 1 }{h} + \: {x}^{x} . \lim_{h \to 0} \frac{ {x }^{ h}- 1}{h}$

$\displaystyle \sf = {x}^{x }. \ln e + \: {x}^{x} . \ln x$

$\displaystyle \sf = {x}^{x } + \: {x}^{x} . \ln x$

$\displaystyle \sf = {x}^{x }(1 + \ln x)$

FINAL ANSWER

Using first principle on derivatives

$\displaystyle \sf \frac{d}{dx} \bigg( {x}^{x} \bigg) = {x}^{x }(1 + \ln x)$

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