The set of all points, where the function f(x)= 1- ^ -x^2 is differentiable, is

Correct answer: C. (-∞, 0) (0, ∞) aligned & f(x)= 1-e^ -x^2 & f^ (x)= 1 2 1-e^ -x^2 ·(-e^ -x^2 )(-2 x) aligned ^ (x)= xe^ -x^2 1- ^ -x^2 ^ (x) is defined for all values of x , except 0 . f(x) is differentiable on (-∞, 0) (0, ∞)

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Continuity and Differentiability

NDA 2017 · Mathematics · Continuity and Differentiability

The set of all points, where the function f(x)= 1- ^ -x^2 is...

The set of all points, where the function

f(\mathrm{x})=\sqrt{1-\mathrme^{-\mathrm{x}^2}}

is differentiable, is

Options

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

Correct answer

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

Step-by-step solution

\begin{aligned} & f(x)=\sqrt{1-e^{-x^2}} \\ & f^{\prime}(x)=\frac{1}{2 \sqrt{1-e^{-x^2}}} \cdot\left(-e^{-x^2}\right)(-2 x)\end{aligned}

\mathrmf^{\prime}(\mathrm{x})=\frac{\mathrm{xe}^{-\mathrm{x}^2}}{\sqrt{1-\mathrme^{-\mathrm{x}^2}}}\mathrmf^{\prime}(\mathrm{x})

is defined for all values of x , except 0 .

\therefore \mathrm{f}(\mathrm{x})

is differentiable on

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

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