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The set of all points where the function \mathrm{f}(\mathrm{x})=\sqrt{1-e^{-x^2}}\mathrm{f}(\mathrm{x})=\sqrt{1-e^{-x^2}} is differentiable is

Option: 1

(0, \infty)


Option: 2

(-\infty, \infty)


Option: 3

(-\infty, \infty) \boxtimes\{0\}


Option: 4

None of these


Answers (1)

best_answer

For \mathrm{x} \neq 0f^{\prime}(x)=\frac{1}{2} \frac{1}{\sqrt{1-e^{-x^2}}}\left[-(-2 x) e^{-x^2}\right]

=\frac{x e^{-x^2}}{\sqrt{1-e^{-x^2}}}

Also, f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}}\left(\frac{e^{-h^2}-1}{-h^2}\right)^{1 / 2}=1And

f^{\prime}\left(0^{-}\right)=-\lim _{h \rightarrow 0^{-}}\left(\frac{e^{-h^2}-1}{-h^2}\right)^{1 / 2}=-1

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Gunjita

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