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