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Question Number 201393 by mokys last updated on 05/Dec/23
by diffention find f ′(z) of f(z) = (z)^(1/3)
$${by}\:{diffention}\:{find}\:{f}\:'\left({z}\right)\:{of}\:{f}\left({z}\right)\:=\:\sqrt[{\mathrm{3}}]{{z}} \\ $$
Commented by mokys last updated on 05/Dec/23
how can solve this
$${how}\:{can}\:{solve}\:{this} \\ $$
Answered by aleks041103 last updated on 05/Dec/23
this is trivial f(z)=(z)^(1/3) =z^(1/3) f ′(z)=lim_(h→0) ((f(z+h)−f(z))/h)= =lim_(h→0) (((z+h)^(1/3) −z^(1/3) )/h)= =lim_(h→0) ((e^(ln(z+h)/3) −e^(ln(z)/3) )/h)= =e^(ln(z)/3) lim_(h→0) ((e^(ln(1+h/z)/3) −1)/h)= =z^(1/3) lim_(h→0) ((e^((h/z+o(h/z))/3) −1)/h)= =z^(1/3) lim_(h→0) (((1+(h/(3z))+o((h/z))+o((h/(3z))+o((h/(3z)))))−1)/h)= =z^(1/3) lim_(h→0) ((1/(3z))+(1/z) ((o(h/z))/(h/z)))= (iff z≠0) =(z^(1/3) /(3z))+(z^(1/3) /z)lim_(g→0) ((o(g))/g)= =(1/3)z^(−2/3) ⇒f ′(z)= { (((1/(3(z^2 )^(1/3) )), z≠0)),((undeff., z=0)) :}
$${this}\:{is}\:{trivial} \\ $$$${f}\left({z}\right)=\sqrt[{\mathrm{3}}]{{z}}={z}^{\mathrm{1}/\mathrm{3}} \\ $$$${f}\:'\left({z}\right)=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left({z}+{h}\right)−{f}\left({z}\right)}{{h}}= \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left({z}+{h}\right)^{\mathrm{1}/\mathrm{3}} −{z}^{\mathrm{1}/\mathrm{3}} }{{h}}= \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{{e}^{{ln}\left({z}+{h}\right)/\mathrm{3}} −{e}^{{ln}\left({z}\right)/\mathrm{3}} }{{h}}= \\ $$$$={e}^{{ln}\left({z}\right)/\mathrm{3}} \underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{{e}^{{ln}\left(\mathrm{1}+{h}/{z}\right)/\mathrm{3}} −\mathrm{1}}{{h}}= \\ $$$$={z}^{\mathrm{1}/\mathrm{3}} \underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{{e}^{\frac{{h}/{z}+{o}\left({h}/{z}\right)}{\mathrm{3}}} −\mathrm{1}}{{h}}= \\ $$$$={z}^{\mathrm{1}/\mathrm{3}} \underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left(\mathrm{1}+\frac{{h}}{\mathrm{3}{z}}+{o}\left(\frac{{h}}{{z}}\right)+{o}\left(\frac{{h}}{\mathrm{3}{z}}+{o}\left(\frac{{h}}{\mathrm{3}{z}}\right)\right)\right)−\mathrm{1}}{{h}}= \\ $$$$={z}^{\mathrm{1}/\mathrm{3}} \underset{{h}\rightarrow\mathrm{0}} {{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{3}{z}}+\frac{\mathrm{1}}{{z}}\:\frac{{o}\left({h}/{z}\right)}{{h}/{z}}\right)= \\ $$$$\left({iff}\:{z}\neq\mathrm{0}\right)\:=\frac{{z}^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}{z}}+\frac{{z}^{\mathrm{1}/\mathrm{3}} }{{z}}\underset{{g}\rightarrow\mathrm{0}} {{lim}}\frac{{o}\left({g}\right)}{{g}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{z}^{−\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{f}\:'\left({z}\right)=\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{{z}^{\mathrm{2}} }},\:{z}\neq\mathrm{0}}\\{{undeff}.,\:{z}=\mathrm{0}}\end{cases} \\ $$$$ \\ $$

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