Which of the following functions is differentiable at x = 0 ?

We have,

$\cos |x|= \begin{cases} \qquad ~~\cos x~~~ , \qquad x \geq 0 \\ \cos (-x)=\cos x, ~~ x<0\end{cases}$

⇒ cos |x| = cos x  for all x ∈ R

Similarly, we have,

$\sin |x|= \begin{cases}\sin x & x \geq 0 \\ \sin (-x)=-\sin x, & x<0\end{cases}$

Let $f(x)=\cos (x)+|x|$ and $g(x)=\cos |x|-|x|$.

Since $\cos x$ is everywhere differentiable and |x| is not differentiable at x = 0. Therefore, f(x) and g(x) are not differentiable at x = 0.

Let $u(x)=\sin (|x|)+|x|$. Then,

$u(x)= \begin{cases}\sin x+x, & x \geq 0 \\ -\sin x-x, & x<0\end{cases}$

∴ (LHD at x = 0) = $\left\{\frac{d}{d x}(-\sin x-x)\right\}_{\text {at } x=0}$

⇒ (LHD at x = 0) = $\{-\cos x-1\}_{\text {at }} x=0=-\cos 0-1=-2$

and,

(RHD at x = 0) = $\left\{\frac{d}{d x}(\sin x+x)\right\}_{\text {at } x=0}=\{\cos x+1\}_{\text {at } x=0}=2$

Clearly, (LHD at x = 0) ≠ (RHD at x = 0)

So, u(x) is not differentiable at x = 0.

Let $v(x)=\sin |x|-|x|$. Then,

$v(x)=\left\{\begin{aligned} -\sin x+x, & x<0 \\ \sin x-x, & x \geq 0 \end{aligned}\right. $

∴ (LHD at x = 0) = $\left\{\frac{d}{d x}(-\sin x+x)\right\}_{\text {at } x=0}$

⇒ (LHD at x = 0) = $\{-\cos x+1\}_{\text {at } x=0}=-1+1=0$

∴ (RHD at x = 0) = $\left\{\frac{d}{d x}(\sin x-x)\right\}_{\text {at } x=0}$

⇒ (RHD at x = 0) = $\{\cos x-1\}_{\text {at }} x=0=\cos 0-1=0$

∴ (LHD at x = 0) = (RHD at x = 0)

So, v(x) is differentiable at x = 0.