Question:
If
\[
f(x)=
\begin{cases}
\tan^{-1}x, & |x|\leq 1\\[2mm]
\frac{1}{2}\left(|x|-1\right), & |x|\gt 1
\end{cases}
\]
then the domain of
\[
\frac{d}{dx}f(x)
\]
is
If
\[
f(x)=
\begin{cases}
\tan^{-1}x, & |x|\leq 1\\[2mm]
\frac{1}{2}\left(|x|-1\right), & |x|\gt 1
\end{cases}
\]
then the domain of
\[
\frac{d}{dx}f(x)
\]
is
Show Hint
For piecewise functions, differentiability at boundary points requires both continuity and equality of left-hand and right-hand derivatives. Always check all three conditions carefully.
Updated On: Jul 18, 2026
- \(\mathbb{R}-\{-1,1\}\)
- \(\mathbb{R}-(-1,1)\)
- \(\mathbb{R}-[-1,1]\)
- \(\mathbb{R}-\{-1\}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Differentiate \(f(x)\) in each interval.
For \[ |x|\leq 1, \] we have \[ f(x)=\tan^{-1}x \] Therefore, \[ f'(x)=\frac{1}{1+x^2} \] which exists for all \[ -1\lt x\lt 1. \] For \[ |x|\gt 1, \] \[ f(x)=\frac{1}{2}(|x|-1) \] When \[ x\gt 1, \] \[ |x|=x \] Hence, \[ f(x)=\frac{x-1}{2} \] and \[ f'(x)=\frac{1}{2}. \] When \[ x\lt -1, \] \[ |x|=-x \] Hence, \[ f(x)=\frac{-x-1}{2} \] and \[ f'(x)=-\frac{1}{2}. \]
Step 2: Check differentiability at \(x=1\).
First check continuity: From the left, \[ f(1)=\tan^{-1}(1)=\frac{\pi}{4}. \] From the right, \[ \lim_{x\to 1^+}\frac{|x|-1}{2} = \lim_{x\to 1^+}\frac{x-1}{2} =0. \] The function is discontinuous at \[ x=1. \] Hence, \[ f'(1) \] does not exist.
Now compute the one-sided derivatives. Left derivative: \[ f'_-(1)=\frac{1}{1+1^2} =\frac{1}{2}. \] Right derivative: \[ f'_+(1)=\frac{1}{2}. \] Although the one-sided derivatives are equal, the function is not continuous at \(x=1\), so \(f'(1)\) does not exist.
Step 3: Check differentiability at \(x=-1\).
From the right, \[ f'_+(-1) = \frac{1}{1+(-1)^2} = \frac{1}{2}. \] From the left, \[ f'_-( -1 ) = -\frac{1}{2}. \] Since \[ f'_+(-1)\neq f'_-( -1 ), \] the derivative does not exist at \[ x=-1. \]
Step 4: Determine the domain of \(f'(x)\).
The derivative exists everywhere except at \[ x=-1. \] Therefore, \[ \text{Domain}(f') = \mathbb{R}-\{-1\}. \]
Step 5: Final conclusion.
Hence, \[ \boxed{\mathbb{R}-\{-1\}} \]
For \[ |x|\leq 1, \] we have \[ f(x)=\tan^{-1}x \] Therefore, \[ f'(x)=\frac{1}{1+x^2} \] which exists for all \[ -1\lt x\lt 1. \] For \[ |x|\gt 1, \] \[ f(x)=\frac{1}{2}(|x|-1) \] When \[ x\gt 1, \] \[ |x|=x \] Hence, \[ f(x)=\frac{x-1}{2} \] and \[ f'(x)=\frac{1}{2}. \] When \[ x\lt -1, \] \[ |x|=-x \] Hence, \[ f(x)=\frac{-x-1}{2} \] and \[ f'(x)=-\frac{1}{2}. \]
Step 2: Check differentiability at \(x=1\).
First check continuity: From the left, \[ f(1)=\tan^{-1}(1)=\frac{\pi}{4}. \] From the right, \[ \lim_{x\to 1^+}\frac{|x|-1}{2} = \lim_{x\to 1^+}\frac{x-1}{2} =0. \] The function is discontinuous at \[ x=1. \] Hence, \[ f'(1) \] does not exist.
Now compute the one-sided derivatives. Left derivative: \[ f'_-(1)=\frac{1}{1+1^2} =\frac{1}{2}. \] Right derivative: \[ f'_+(1)=\frac{1}{2}. \] Although the one-sided derivatives are equal, the function is not continuous at \(x=1\), so \(f'(1)\) does not exist.
Step 3: Check differentiability at \(x=-1\).
From the right, \[ f'_+(-1) = \frac{1}{1+(-1)^2} = \frac{1}{2}. \] From the left, \[ f'_-( -1 ) = -\frac{1}{2}. \] Since \[ f'_+(-1)\neq f'_-( -1 ), \] the derivative does not exist at \[ x=-1. \]
Step 4: Determine the domain of \(f'(x)\).
The derivative exists everywhere except at \[ x=-1. \] Therefore, \[ \text{Domain}(f') = \mathbb{R}-\{-1\}. \]
Step 5: Final conclusion.
Hence, \[ \boxed{\mathbb{R}-\{-1\}} \]
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