Question

Which of the following function(s) is differentiable at x = 0?
(a) \[\cos \left( \left| x \right| \right)+\left| x \right|\]
(b) \[\cos \left( \left| x \right| \right)-\left| x \right|\]
(c) \[\sin \left( \left| x \right| \right)+\left| x \right|\]
(d) \[\sin \left( \left| x \right| \right)-\left| x \right|\]

Answer 1

Hint: Differentiate the given function and check \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\]= finite quantity for differentiability at x = 0.

Here we have to check if the function given is differentiable at x = 0 or not.
We know that for the function to be differentiable at x = 0, it must satisfy \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\]= finite quantity\[....\left( i \right)\]
We know that \[\left| x \right|=x\text{ for }x\ge 0....\left( ii \right)\]
And, \[\left| x \right|=-x\text{ for }x<0....\left( iii \right)\]
Taking \[f\left( x \right)=\cos \left( \left| x \right| \right)+\left| x \right|\]
By equation (ii) and (iii)
\[f\left( x \right)=\left\{ \begin{align}
  & \cos \left( x \right)+x,\text{ }x\ge 0 \\
 & \cos \left( -x \right)-x,\text{ }x<0 \\
\end{align} \right.\]
As we know that \[\cos \left( -x \right)=\cos x\]
We get, \[f\left( x \right)=\left\{ \begin{align}
  & \cos x+x,\text{ }x\ge 0 \\
 & \cos x-x,\text{ }x<0 \\
\end{align} \right.\]
Now, we will differentiate \[f\left( x \right)\]
Since, we know that \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]and \[\dfrac{d}{dx}\left( x \right)=1\]
We get, \[{{f}^{'}}\left( x \right)=\left\{ \begin{align}
  & -\sin x+1\text{, }x>0 \\
 & -\sin x-1,\text{ x0} \\
\end{align} \right.\]
Now, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\sin x-1 \right)=-\sin \left( {{0}^{o}} \right)-1\] [Since \[\sin \left( {{0}^{o}} \right)=0\]]
We get \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=-1\]
Now, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -\sin x+1 \right)=-\sin \left( {{0}^{o}} \right)+1=1\]
Since, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\]
Therefore, \[f\left( x \right)=\cos \left( \left| x \right| \right)+\left| x \right|\]is not differentiable at x = 0.
Now taking \[g\left( x \right)=\cos \left( \left| x \right| \right)-\left| x \right|\]
By equation (ii) and (iii)
We get, \[g\left( x \right)=\left\{ \begin{align}
  & cos x-x,\text{ }x\ge 0 \\
 & \cos x+x,\text{ }x<0 \\
\end{align} \right.\]
Now, we will differentiate \[g\left( x \right)\].
We get, \[{{g}^{'}}\left( x \right)=\left\{ \begin{align}
  & -\sin x-1,\text{ }x>0 \\
 & -\sin x+1,\text{ }x<0 \\
\end{align} \right.\text{ }\!\![\!\!\text{ Also }\sin \left( {{0}^{o}} \right)\text{=0 }\!\!]\!\!\text{ }\]
Now, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\sin x+1 \right)=-\sin \left( {{0}^{o}} \right)+1=1\]
And \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -\sin x-1 \right)=-\sin \left( {{0}^{o}} \right)-1=-1\]
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)\]
Therefore, \[g\left( x \right)=\cos \left( \left| x \right| \right)-\left| x \right|\]is not differentiable at x = 0.
Now, taking \[h\left( x \right)=\sin \left( \left| x \right| \right)+\left| x \right|\]
By equation (ii) and (iii),
We get \[h\left( x \right)=\left\{ \begin{align}
  & \sin x+x,\text{ }x\ge 0 \\
 & \sin \left( -x \right)-x,\text{ }x<0 \\
\end{align} \right.\text{ }\!\![\!\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\!\!]\!\!\text{ }\]
\[h\left( x \right)=\left\{ \begin{align}
  & \sin x+x,\text{ }x\ge 0 \\
 & -\sin x-x,\text{ }x<0 \\
\end{align} \right.\]
Now, we will differentiate \[h\left( x \right)\].
Since, \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
We get, \[{{h}^{'}}\left( x \right)=\left\{ \begin{align}
  & \cos x+1,\text{ }x>0 \\
 & -\cos x-1,\text{ }x<0 \\
\end{align} \right.\text{ }\!\![\!\!\text{ Also }\cos \left( {{0}^{o}} \right)\text{=1 }\!\!]\!\!\text{ }\]
Now, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\cos x-1 \right)=-\cos \left( {{0}^{o}} \right)-1=-1-1=-2\]
\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \cos x+1 \right)=\cos \left( {{0}^{o}} \right)+1=1+1=2\]
Since, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)\]
Therefore, \[h\left( x \right)=\sin \left( \left| x \right| \right)+\left| x \right|\]is not differentiable at x = 0.
Now, taking \[J\left( x \right)=\sin \left( \left| x \right| \right)-\left| x \right|\]
By equation (ii) and (iii),
We get \[J\left( x \right)=\left\{ \begin{align}
  & \sin x-x,\text{ }x\ge 0 \\
 & \sin \left( -x \right)+x,\text{ }x<0 \\
\end{align} \right.\text{ }\!\![\!\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\!\!]\!\!\text{ }\]
\[J\left( x \right)=\left\{ \begin{align}
  & \sin x-x,\text{ }x\ge 0 \\
 & -\sin x+x,\text{ }x<0 \\
\end{align} \right.\]
Now, we will differentiate \[J\left( x \right)\].
Since, \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
We get, \[{{J}^{'}}\left( x \right)=\left\{ \begin{align}
  & \cos x-1,\text{ }x>0 \\
 & -\cos x+1,\text{ }x<0 \\
\end{align} \right.\text{ }\!\![\!\!\text{ Also }\cos \left( {{0}^{o}} \right)\text{=1 }\!\!]\!\!\text{ }\]
Now, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \cos x-1 \right)=\cos \left( {{0}^{o}} \right)-1=1-1=0\]
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\cos x+1 \right)=-\cos \left( {{0}^{o}} \right)+1=-1+1=0\]
Since, \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)\]
Therefore, J(x) is differentiable at x = 0.
Hence, option (d) is correct.

Note: Some students check the continuity first in these types of questions but that is not required. It makes the solution time consuming, because if a function is differentiable, it would surely be continuous as well and need not be checked.