Question

Finch Derivative of log x by using
first prinupal Mithred​

Answer 1

$\underline\blue{\bold{Given \: Question :- }}$

• Find the Derivative of log x by using first principal method

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$\huge \orange{AηsωeR} ✍$

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${ \boxed {\bf{Given}}}$

• f(x) = logx

${ \boxed {\bf{To \: Find}}}$

• f'(x) using first principal.

${ \boxed {\bf{Formula \: used :- }}}$

$\bf \:\lim_{x\rightarrow 0}  log(\dfrac{1 + x}{x} ) = 1$

$\bf \: log(a) - log(b) = log(\dfrac{a}{b} )$

${ \boxed {\bf{Solution}}}$

$\bf \:Let \: f(x) = log(x)$

☆Change x to x + h

$\bf\implies \:f(x + h) = log(x + h)$

☆By definition of First Principal of Differentiation,

$\bf \:f'(x) = \lim_{h\rightarrow 0} \dfrac{f(x + h) - f(x)}{h}$

$\bf\implies \:f'(x) = \lim_{h\rightarrow 0} \dfrac{ log(x + h) - log(x) }{h}$

$\bf\implies \:f'(x) = \lim_{h\rightarrow 0} \dfrac{ log(\dfrac{x + h}{x} ) }{h}$

$\bf\implies \:f'(x) = \lim_{h\rightarrow 0} \dfrac{ log(1 + \dfrac{h}{x} ) }{h}$

☆Multiply and divide by x the denominator, we get

$\bf\implies \:f'(x) = \lim_{h\rightarrow 0} \dfrac{ log(1 + \dfrac{h}{x} ) }{\dfrac{h}{x} \times x }$

$\bf\implies \:f'(x) =\dfrac{1}{x} \lim_{h\rightarrow 0} \dfrac{ log(1 + \dfrac{h}{x} ) }{\dfrac{h}{x} }$

$\bf\implies \:f'(x) = \dfrac{1}{x} \times 1 = \dfrac{1}{x}$

$\large{\boxed{\boxed{\bf{Hence, \dfrac{d}{dx} logx = \dfrac{1}{x} }}}}$