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Check if the function $f(x)=1-(x-1)^{2/3}$ is differentiable . I have used basic limit definition of differentiation, but could not solve the limit.

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  • $\begingroup$ Is $x^{2/3}$ at $0$ ? $\endgroup$ – dmtri Sep 12 '18 at 13:16
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We have that by definition

$$\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}=\lim_{x\to 1^+} \frac{-(x-1)^{2/3}}{x-1}=\lim_{x\to 1^+} -\frac1{(x-1)^{1/3}}\to -\infty$$

therefore $f(x)$ is not differentiable at $x=1$.

Or as an alternative

$$f(x)=1-(x-1)^{2/3} \implies f'(x)=-\frac2{3\sqrt[3]{x-1}}$$

and

$$\lim_{x\to 1^+} f'(x)=-\infty$$

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