Question

p(x)=x^x, then the value of f'(x)​

Answer 1

The correct Statement is

$\rm \: If \: f(x) = {x}^{x} \: then \: the \: value \: of \: f'(x)$

Formula Used :-

$(1). \: \boxed{ \blue{ \: \rm : \implies \:\dfrac{d}{dx} x = 1}}$

$(2). \: \boxed{ \blue{ \: \rm : \implies \: \: \dfrac{d}{dx} log(x) = \dfrac{1}{x} }}$

$(3). \: \boxed{ \blue{ \: \rm : \implies \dfrac{d}{dx} u.v \: = u \: \dfrac{d}{dx} v \: + \: v \: \dfrac{d}{dx} u}}$

$\large\underline\purple{\bold{Solution :- }}$

Given

$\rm : \implies \:f(x) \: = \: {x}^{x}$

On taking log on both sides, we get

$\rm : \implies \:log \: f(x) \: = \: log( {x}^{x} )$

$\rm : \implies \:log \: f(x) = x \: logx$

On differentiating both sides w. r. t. x, we get

$\rm : \implies \: \dfrac{d}{dx} log \: f(x) = \dfrac{d}{dx} (x \: logx)$

$\rm : \implies \:\dfrac{1}{f(x)} \dfrac{d}{dx} f(x) \: = \: x \dfrac{d}{dx} logx \: + \: logx \: \dfrac{d}{dx} x$

$\rm : \implies \:\dfrac{f'(x)}{f(x)} \: = \: x \times \dfrac{1}{x} + logx \: \times 1$

$\rm : \implies \:f'(x) \: = \: f(x) \: (1 + logx)$

$\boxed{ \pink{ \: \rm : \implies \:f'(x) \: = \: {x}^{x} \: (1 + logx)}}$