Differentiation Questions

Question Number 103998 by  M±th+et+s last updated on 18/Jul/20

$${if}\:\:\:\:{f}\left({x}\right)={x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\$$$${f}'\left(\mathrm{0}\right)=\mathrm{0}\:\:{or}\:{not}\:{exist} \\$$

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

After reviewing some books I found this

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

$${i}\:{posted}\:{this}\:{question}\:{from}\:{this}\:{exam}\: \\$$$$\\$$$${and}\:{thanks}\:{for}\:{all} \\$$

Answered by frc2crc last updated on 18/Jul/20

$$\frac{{d}}{{dx}}{x}^{\mathrm{3}/\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}/\mathrm{2}−\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{x}} \\$$$${so}\:{f}'\left(\mathrm{0}\right)=\mathrm{0} \\$$

Commented by  M±th+et+s last updated on 18/Jul/20

$${thank}\:{you}\:{sir}\: \\$$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}f}\left({x}\right)=\mathrm{0}\:\:\:{but}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {{lim}f}\left({x}\right)\:{is}\:{not}\:{exist} \\$$$${so}\:{i}\:{think}\:{we}\:{cant}\:{find}\:{f}'\left(\mathrm{0}\right)\: \\$$

Commented by frc2crc last updated on 18/Jul/20

$${it}\:{doesn}'{t}\:{make}\:{any}\:{sense} \\$$$$\sqrt{{x}\:}{is}\:'{only}'\:{for}\:{positive} \\$$$${numbers},{why}\:{do}\:{you}\:{need} \\$$$${a}\:{limit}? \\$$$$\\$$

Commented by  M±th+et+s last updated on 18/Jul/20

$${i}\:{asked}\:{about}\:{the}\:{limit}\:{because} \\$$$${f}'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{{f}\left({x}\right)−{f}\left({a}\right)}{{x}−{a}} \\$$$${f}'\left(\mathrm{0}\right)=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\sqrt{{x}}−\mathrm{0}}{{x}−\mathrm{0}}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\sqrt{{x}} \\$$$$\\$$$${and}\:{i}\:{wasn}^{,} {t}\:{sure}\:{if}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\sqrt{{x}}=\mathrm{0} \\$$$$\\$$$${and}\:{thank}\:{you}\:{for}\:{your}\:{explaining} \\$$$$\\$$

Answered by OlafThorendsen last updated on 18/Jul/20

$$\frac{{f}\left(\mathrm{0}+{h}\right)−{f}\left(\mathrm{0}\right)}{{h}}\:=\:\frac{{h}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{0}}{{h}}\:=\:\sqrt{{h}} \\$$$$\underset{{h}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{f}\left(\mathrm{0}+{h}\right)−{f}\left(\mathrm{0}\right)}{{h}}\:=\:\mathrm{0} \\$$$$\underset{{h}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{{f}\left(\mathrm{0}+{h}\right)−{f}\left(\mathrm{0}\right)}{{h}}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\$$$$\\$$

Commented by  M±th+et+s last updated on 18/Jul/20

$${yes}\:{sir}\:{that}\:{what}\:{was}\:{i}\:{mean} \\$$

Commented by bubugne last updated on 18/Jul/20

$${with}\:{k}\:=\:−{h} \\$$$${h}\:=\:{i}^{\mathrm{2}} {k}\:\left({k}\geqslant\mathrm{0}\right) \\$$$$\underset{{h}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\sqrt{{h}}\:=\:\underset{{k}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{i}\:\sqrt{{k}}\:=\:{i}\:×\:\mathrm{0}\:=\:\mathrm{0} \\$$

Commented by prakash jain last updated on 18/Jul/20

$$\mathrm{If}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has} \\$$$$\mathrm{left}\:\mathrm{end}\:\mathrm{point}\:\mathrm{then}\:\mathrm{you}\:\mathrm{only}\:\mathrm{need} \\$$$$\mathrm{RHL}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{continuity}.\:\mathrm{You} \\$$$$\mathrm{do}\:\mathrm{not}\:\mathrm{go}\:\mathrm{outside}\:\mathrm{of}\:\mathrm{domain}\:\mathrm{and} \\$$$$\mathrm{say}\:\mathrm{function}\:\mathrm{is}\:\mathrm{discontinuous}. \\$$$${f}\left({x}\right)=\sqrt{{x}}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:{x}=\mathrm{0}. \\$$

Answered by Ar Brandon last updated on 18/Jul/20

$$\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{differentiable}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{x}_{\mathrm{0}} \:\mathrm{iff} \\$$$$\underset{\mathrm{x}\rightarrow\mathrm{x}_{\mathrm{0}} } {\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)}{\mathrm{x}−\mathrm{x}_{\mathrm{0}} }\:\mathrm{exists}. \\$$$$\underset{\mathrm{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{0}}{\mathrm{x}−\mathrm{0}}=\mathrm{0}\:,\:\underset{\mathrm{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{0}}{\mathrm{x}−\mathrm{0}}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}. \\$$$$\mathrm{Therefore}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{differentiable}\:\mathrm{at}\:\mathrm{x}=\mathrm{0} \\$$

Commented by prakash jain last updated on 18/Jul/20

$$\\$$$$\mathrm{If}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has} \\$$$$\mathrm{left}\:\mathrm{end}\:\mathrm{point}\:\mathrm{then}\:\mathrm{you}\:\mathrm{only}\:\mathrm{need} \\$$$$\mathrm{RHL}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{continuity}.\:\mathrm{You} \\$$$$\mathrm{do}\:\mathrm{not}\:\mathrm{go}\:\mathrm{outside}\:\mathrm{of}\:\mathrm{domain}\:\mathrm{and} \\$$$$\mathrm{say}\:\mathrm{function}\:\mathrm{is}\:\mathrm{discontinuous}. \\$$$${f}\left({x}\right)=\sqrt{{x}}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:{x}=\mathrm{0}. \\$$$$\mathrm{It}\:\mathrm{is}\:\mathrm{incorrect}\:\mathrm{to}\:\mathrm{take}\:\mathrm{LHL}\:\mathrm{for} \\$$$$\mathrm{function}\:\sqrt{{x}} \\$$

Commented by Ar Brandon last updated on 19/Jul/20

OK Sir. Thanks for the clarification.