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Question Number 169366 by Mastermind last updated on 29/Apr/22
Differentiate from first principle  y=log_x a    Mastermind
$${Differentiate}\:{from}\:{first}\:{principle} \\ $$$${y}={log}_{{x}} {a} \\ $$$$ \\ $$$${Mastermind} \\ $$
Commented by bobhans last updated on 29/Apr/22
  y = ((ln a)/(ln x)) ⇒(dy/dx) = lim_(h→0)  ((((ln a)/(ln (x+h))) − ((ln a)/(ln x)))/h)   (dy/dx) = ln a lim_(h→0)  ((ln x−ln (x+h))/(h ln x ln (x+h)))   (dy/dx) = −((ln a)/(ln^2 (x))). lim_(h→0)  ((ln (((x+h)/x)))/h)           =−((ln a)/(ln^2 (x))) .lim_(h→0)  ((ln (1+(h/x)))/h)           =−((ln a)/(x ln^2 (x)))
$$\:\:{y}\:=\:\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{x}}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:\left({x}+{h}\right)}\:−\:\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{x}}}{{h}} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\mathrm{ln}\:{a}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:{x}−\mathrm{ln}\:\left({x}+{h}\right)}{{h}\:\mathrm{ln}\:{x}\:\mathrm{ln}\:\left({x}+{h}\right)} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:−\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)}.\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\frac{{x}+{h}}{{x}}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)}\:.\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{ln}\:{a}}{{x}\:\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)}\: \\ $$

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