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)

& \cos \left( x \right)+x,\text{ }x\ge 0 \\\ & \cos \left( -x \right)-x,\text{ }x<0 \\\ \end{aligned} \right.$$ As we know that $$\cos \left( -x \right)=\cos x$$ We get, $$f\left( x \right)=\left\\{ \begin{aligned} & \cos x+x,\text{ }x\ge 0 \\\ & \cos x-x,\text{ }x<0 \\\ \end{aligned} \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{aligned} & -\sin x+1\text{, }x>0 \\\ & -\sin x-1,\text{ x0} \\\ \end{aligned} \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{aligned} & cos x-x,\text{ }x\ge 0 \\\ & \cos x+x,\text{ }x<0 \\\ \end{aligned} \right.$$ Now, we will differentiate $$g\left( x \right)$$. We get, $${{g}^{'}}\left( x \right)=\left\\{ \begin{aligned} & -\sin x-1,\text{ }x>0 \\\ & -\sin x+1,\text{ }x<0 \\\ \end{aligned} \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{aligned} & \sin x+x,\text{ }x\ge 0 \\\ & \sin \left( -x \right)-x,\text{ }x<0 \\\ \end{aligned} \right.\text{ }\\!\\![\\!\\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\\!\\!]\\!\\!\text{ }$$ $$h\left( x \right)=\left\\{ \begin{aligned} & \sin x+x,\text{ }x\ge 0 \\\ & -\sin x-x,\text{ }x<0 \\\ \end{aligned} \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{aligned} & \cos x+1,\text{ }x>0 \\\ & -\cos x-1,\text{ }x<0 \\\ \end{aligned} \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{aligned} & \sin x-x,\text{ }x\ge 0 \\\ & \sin \left( -x \right)+x,\text{ }x<0 \\\ \end{aligned} \right.\text{ }\\!\\![\\!\\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\\!\\!]\\!\\!\text{ }$$ $$J\left( x \right)=\left\\{ \begin{aligned} & \sin x-x,\text{ }x\ge 0 \\\ & -\sin x+x,\text{ }x<0 \\\ \end{aligned} \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{aligned} & \cos x-1,\text{ }x>0 \\\ & -\cos x+1,\text{ }x<0 \\\ \end{aligned} \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.