Question

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

• A. $(0,\infty )$
• B. $(-\infty ,\infty )$
• C. $(-\infty ,0)\cup (0,\infty )$
• D. $(-1,\infty )$

Answer 1

Answer: (C)

$(-\infty ,0)\cup (0,\infty )$

Given that

$f\left(x\right)=\sqrt{1-e^{-x^2}}$

By chain rule of differentiation,

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{1-e^{-x^2}}}\frac{d}{dx}(1-e^{-x^2})$

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

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

f(x) is differentiable at a point x if the f'(x) exists at the point.

The denominator of f'(x) vanishes for

$1-e^{-x^2}=0$

i.e for

$x=0$

f(x) is not differentiable at x = 0.

Hence, the set of all points at which f(x) is differentiable:

$(-\infty ,0)\cup (0,\infty )$

The answer: option C.