Question Number 217664 by hardmath last updated on 17/Mar/25
$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}}}\: \\ $$
Answered by SdC355 last updated on 18/Mar/25
$${f}^{\left(\mathrm{1}\right)} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}. \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\sqrt{\mathrm{2}{x}+\mathrm{3}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}\left({x}+{h}\right)+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}}{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}}{{h}}= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}{{h}}\centerdot\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}{\:\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)} \\ $$$$\therefore\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}−\left(\mathrm{2}{x}+\mathrm{3}\right)}{{h}\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\cancel{\mathrm{2}{x}}+\mathrm{2}{h}+\cancel{\mathrm{3}}−\cancel{\mathrm{2}{x}}−\cancel{\mathrm{3}}}{{h}\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\cancel{{h}}}{\cancel{{h}}\left(\sqrt{\mathrm{2}{x}+{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right.}=\frac{\cancel{\mathrm{2}}}{\cancel{\mathrm{2}}\sqrt{\mathrm{2}{x}+\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{3}}} \\ $$
Commented by hardmath last updated on 21/Mar/25
$$\mathrm{thankyou}\:\mathrm{dearSir} \\ $$