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Question Number 205013 by mathlove last updated on 05/Mar/24

$$ify=\sqrt[7]{x}provethat$$$$y^{{}^{\prime }}=\frac{1}{7\sqrt[7]{x^6}}$$

Answered by Frix last updated on 05/Mar/24

$$y=x^r\Rightarrow y^{\prime }={rx}^{r-1};r\ne 0$$$$y=\sqrt[7]{x}=x^{\frac{1}{7}}\Rightarrow y^{\prime }=\frac{1}{7}{x}^{-\frac{6}{7}}=\frac{1}{7\sqrt[7]{x^6}}$$

Answered by MM42 last updated on 06/Mar/24

$$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\sqrt[7]{x+h}-\sqrt[7]{x}}{h}$$$$={lim}_{h\to 0}\frac{h}{h(\sqrt[7]{{(x+h)}^6}+\sqrt[7]{{(x+h)}^{5}x}+...+\sqrt[7]{x^6})}$$$$=\frac{1}{7\sqrt[7]{x^6}}✓$$$$$$

Commented by mathlove last updated on 06/Mar/24

$$thanks$$

Commented by TonyCWX08 last updated on 06/Mar/24

$$Wrong!$$$$Derivativeof\sqrt[7]{x}$$$$=\frac{1}{7\sqrt[7]{x^6}}$$

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