Differentiate $ \displaystyle y=\sqrt[3]{x}$ from the first principles.
Solution
$ \displaystyle \ \ \ y=\sqrt[3]{x}={{x}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore y+\delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore \delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}\times \frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$ (***)
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{x+\delta x-x}}{{\delta x\left( {{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}} \right)}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \therefore \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{{{x}^{{\frac{2}{3}}}}+{{x}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{3{{x}^{{\frac{2}{3}}}}}}$
Solution αဲ့ line 5 (***) αွာ indeterminate form αို ေျαာα္းαိုα conjugate αဲα ေျαႇာα္αာ ျαα ္αါαα္။
Solution
$ \displaystyle \ \ \ y=\sqrt[3]{x}={{x}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore y+\delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore \delta y={{\left( {x+\delta x} \right)}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}-{{x}^{{\frac{1}{3}}}}}}{{\delta x}}\times \frac{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$ (***)
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{{x+\delta x-x}}{{\delta x\left( {{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}} \right)}}$
$ \displaystyle \therefore \frac{{\delta y}}{{\delta x}}=\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \therefore \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{1}{{{{{\left( {x+\delta x} \right)}}^{{\frac{2}{3}}}}+{{{\left( {x+\delta x} \right)}}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{{{x}^{{\frac{2}{3}}}}+{{x}^{{\frac{1}{3}}}}{{x}^{{\frac{1}{3}}}}+{{x}^{{\frac{2}{3}}}}}}$
$ \displaystyle \ \ \ \ \ \ \ \,=\frac{1}{{3{{x}^{{\frac{2}{3}}}}}}$
Solution αဲ့ line 5 (***) αွာ indeterminate form αို ေျαာα္းαိုα conjugate αဲα ေျαႇာα္αာ ျαα ္αါαα္။
$ \displaystyle {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$
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