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Question Number 217664 by hardmath last updated on 17/Mar/25

If f(x) = (√(2x + 3)) prove that: f ′(x) = (1/( (√(2x + 3))))

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}}}\: \\ $$

Answered by deleted50 last updated on 18/Mar/25

f^((1)) (x)=lim_(h→0) ((f(x+h)−f(x))/h). ((d )/dx)(√(2x+3))=lim_(h→0) (((√(2(x+h)+3))−(√(2x+3)))/h) lim_(h→0) (((√(2x+2h+3))−(√(2x+3)))/h)= lim_(h→0) ((((√(2x+2h+3))−(√(2x+3))))/h)∙((((√(2x+2h+3))+(√(2x+3))))/( ((√(2x+2h+3))+(√(2x+3))))) ∴lim_(h→0) ((2x+2h+3−(2x+3))/(h((√(2x+2h+3))+(√(2x+3))))) lim_(h→0) ((2x+2h+3−2x−3)/(h((√(2x+2h+3))+(√(2x+3)))))= lim_(h→0) ((2h)/(h((√(2x+h+3))+(√(2x+3))))=(2/(2(√(2x+3)))) (1/( (√(2x+3))))

$${f}^{\left(\mathrm{1}\right)} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}}. \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\sqrt{\mathrm{2}{x}+\mathrm{3}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}\left({x}+{h}\right)+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}}{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}}{{h}}= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}−\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}{{h}}\centerdot\frac{\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}{\:\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)} \\ $$$$\therefore\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}−\left(\mathrm{2}{x}+\mathrm{3}\right)}{{h}\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\cancel{\mathrm{2}{x}}+\mathrm{2}{h}+\cancel{\mathrm{3}}−\cancel{\mathrm{2}{x}}−\cancel{\mathrm{3}}}{{h}\left(\sqrt{\mathrm{2}{x}+\mathrm{2}{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right)}= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\cancel{{h}}}{\cancel{{h}}\left(\sqrt{\mathrm{2}{x}+{h}+\mathrm{3}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}\right.}=\frac{\cancel{\mathrm{2}}}{\cancel{\mathrm{2}}\sqrt{\mathrm{2}{x}+\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{3}}} \\ $$

Commented by hardmath last updated on 21/Mar/25

thankyou dearSir

$$\mathrm{thankyou}\:\mathrm{dearSir} \\ $$

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