Question Number 205013 by mathlove last updated on 05/Mar/24
$${if}\:{y}=\sqrt[{\mathrm{7}}]{{x}}\:{prove}\:{that} \\ $$$${y}^{'} =\frac{\mathrm{1}}{\mathrm{7}\:\sqrt[{\mathrm{7}}]{{x}^{\mathrm{6}} }} \\ $$
Answered by Frix last updated on 05/Mar/24
$${y}={x}^{{r}} \:\Rightarrow\:{y}'={rx}^{{r}−\mathrm{1}} ;\:{r}\neq\mathrm{0} \\ $$$${y}=\sqrt[{\mathrm{7}}]{{x}}={x}^{\frac{\mathrm{1}}{\mathrm{7}}} \:\Rightarrow\:{y}'=\frac{\mathrm{1}}{\mathrm{7}}{x}^{−\frac{\mathrm{6}}{\mathrm{7}}} =\frac{\mathrm{1}}{\mathrm{7}\sqrt[{\mathrm{7}}]{{x}^{\mathrm{6}} }} \\ $$
Answered by MM42 last updated on 06/Mar/24
$${f}'\left({x}\right)={lim}_{{h}\rightarrow\mathrm{0}} \frac{\sqrt[{\mathrm{7}}]{{x}+{h}}−\sqrt[{\mathrm{7}}]{{x}}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \frac{{h}}{{h}\left(\sqrt[{\mathrm{7}}]{\left({x}+{h}\right)^{\mathrm{6}} }+\sqrt[{\mathrm{7}}]{\left({x}+{h}\right)^{\mathrm{5}} {x}}+…+\sqrt[{\mathrm{7}}]{{x}^{\mathrm{6}} }\right)} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{7}\sqrt[{\mathrm{7}}]{{x}^{\mathrm{6}} }}\:\:\checkmark \\ $$$$ \\ $$
Commented by mathlove last updated on 06/Mar/24
$${thanks} \\ $$
Commented by TonyCWX08 last updated on 06/Mar/24
$${Wrong}! \\ $$$${Derivative}\:{of}\:\sqrt[{\mathrm{7}}]{{x}\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}\sqrt[{\mathrm{7}}]{{x}^{\mathrm{6}} }} \\ $$