by diffention find f ′(z) of f(z) = (z)^(1/3)


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Question Number 201393 by mokys last updated on 05/Dec/23

$b y d i f f e n t i o n f i n d f^{'} (z) o f f (z) = \sqrt[3]{z}$
Commented by mokys last updated on 05/Dec/23

$h o w c a n s o l v e t h i s$
	Answered by aleks041103 last updated on 05/Dec/23
	$this is trivial f(z)=(z)^(1/3) =z^(1/3) f ′(z)=lim_(h→0) ((f(z+h)−f(z))/h)= =lim_(h→0) (((z+h)^(1/3) −z^(1/3) )/h)= =lim_(h→0) ((e^(ln(z+h)/3) −e^(ln(z)/3) )/h)= =e^(ln(z)/3) lim_(h→0) ((e^(ln(1+h/z)/3) −1)/h)= =z^(1/3) lim_(h→0) ((e^((h/z+o(h/z))/3) −1)/h)= =z^(1/3) lim_(h→0) (((1+(h/(3z))+o((h/z))+o((h/(3z))+o((h/(3z)))))−1)/h)= =z^(1/3) lim_(h→0) ((1/(3z))+(1/z) ((o(h/z))/(h/z)))= (iff z≠0) =(z^(1/3) /(3z))+(z^(1/3) /z)lim_(g→0) ((o(g))/g)= =(1/3)z^(−2/3) ⇒f ′(z)= { (((1/(3(z^2 )^(1/3) )), z≠0)),((undeff., z=0)) :}$
	$t h i s i s t r i v i a l$ $f (z) = \sqrt[3]{z} = z^{1 / 3}$ $f^{'} (z) = \underset{h \to 0}{l i m} \frac{f (z + h) - f (z)}{h} =$ $= \underset{h \to 0}{l i m} \frac{{(z + h)}^{1 / 3} - z^{1 / 3}}{h} =$ $= \underset{h \to 0}{l i m} \frac{e^{l n (z + h) / 3} - e^{l n (z) / 3}}{h} =$ $= e^{l n (z) / 3} \underset{h \to 0}{l i m} \frac{e^{l n (1 + h / z) / 3} - 1}{h} =$ $= z^{1 / 3} \underset{h \to 0}{l i m} \frac{e^{\frac{h / z + o (h / z)}{3}} - 1}{h} =$ $= z^{1 / 3} \underset{h \to 0}{l i m} \frac{(1 + \frac{h}{3 z} + o (\frac{h}{z}) + o (\frac{h}{3 z} + o (\frac{h}{3 z}))) - 1}{h} =$ $= z^{1 / 3} \underset{h \to 0}{l i m} (\frac{1}{3 z} + \frac{1}{z} \frac{o (h / z)}{h / z}) =$ $(i f f z \neq 0) = \frac{z^{1 / 3}}{3 z} + \frac{z^{1 / 3}}{z} \underset{g \to 0}{l i m} \frac{o (g)}{g} =$ $= \frac{1}{3} z^{- 2 / 3}$ $\Rightarrow f^{'} (z) = {\begin{cases} \frac{1}{3 \sqrt[3]{z^{2}}}, z \neq 0 \\ u n d e f f ., z = 0 \end{cases}$
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