Question

Which of the following function is differentiable at $x = 0?$
A) \[{\cos ^{}}(|x|) + |x|\]
B) \[{\cos ^{}}(|x|) - |x|\]
C) \[{\sin ^{}}(|x|) + |x|\]
D) \[{\sin ^{}}(|x|) - |x|\]

Answer 1

Hint: In this question we have to find which functions is differentiable at $x = 0$. A function is differentiable at $x = 0$, if and only if left hand derivative and right derivative is equal at $x = 0$.

Complete step by step solution: F(x) is differentiable at $x = a \Rightarrow Lf'(a) = Rf'(a)$, where $Lf'(a)$ be left hand derivative and $Rf'(a)$be right hand derivative.
If $Lf'(a) \ne R'f(a),the{n^{}}f{(x)^{}}$is not differentiable at $x = a$.
First of all, check for option A. Here \[f(x) = {\cos ^{}}(|x|) + |x|\] and hence find the differentiability of the function as follow:
If $x < 0$ than
\[\begin{gathered}
   \Rightarrow f(x) = {\cos ^{}}(|x|) + |x| \\
   \Rightarrow f(x) = {\cos ^{}}( - x) + ( - x) \\
\end{gathered} \]
\[ \Rightarrow f(x) = {\cos ^{}}x - x\], as \[\cos ( - x) = \cos x\]
Now, $Lf'(x) = - \sin x - 1$, as $\dfrac{{d(\cos (x))}}{{dx}} = - \sin x$
$\begin{gathered}
   \Rightarrow Lf'(0) = - \sin 0 - 1 \\
   \Rightarrow Lf'(0) = - 1 \\
\end{gathered} $ , $\sin 0 = 0$
If $x > 0$ than
\[\begin{gathered}
   \Rightarrow f(x) = {\cos ^{}}(|x|) + |x| \\
   \Rightarrow f(x) = {\cos ^{}}(x) + x \\
\end{gathered} \]
\[ \Rightarrow f(x) = {\cos ^{}}x + x\]
Now, $Rf'(x) = - \sin x + 1$, as $\dfrac{{d(\cos (x))}}{{dx}} = - \sin x$
$\begin{gathered}
   \Rightarrow Rf'(0) = \sin 0 + 1 \\
   \Rightarrow Rf'(0) = 1 \\
\end{gathered} $ , $\sin 0 = 0$
$ \Rightarrow Lf'(0) \ne Rf'(0)$
Hence this function is not differentiable at x=0. Option A is incorrect
Now, check the differentiability of \[f(x) = {\cos ^{}}(|x|) - |x|\] at x=0 as follow:
If $x < 0$ than
\[\begin{gathered}
   \Rightarrow f(x) = {\cos ^{}}(|x|) - |x| \\
   \Rightarrow f(x) = {\cos ^{}}( - x) - ( - x) \\
\end{gathered} \]
\[ \Rightarrow f(x) = {\cos ^{}}x + x\], as \[\cos ( - x) = \cos x\]
Now, $Lf'(x) = - \sin x + 1$, as $\dfrac{{d(\cos (x))}}{{dx}} = - \sin x$
$ \Rightarrow Lf'(0) = 1$ , $\sin 0 = 0$
If $x > 0$ than
\[ \Rightarrow f(x) = {\cos ^{}}(|x|) - |x|\]
\[ \Rightarrow f(x) = {\cos ^{}}x - x\]
Now, $Rf'(x) = - \sin x - 1$, as $\dfrac{{d(\cos (x))}}{{dx}} = - \sin x$
$ \Rightarrow Rf'(0) = 1$ , $\sin 0 = 0$
$ \Rightarrow Lf'(0) \ne Rf'(0)$
Hence this function is not differentiable at x=0. Option B is incorrect
Now, check the differentiability of \[f(x) = {\sin ^{}}(|x|) + |x|\] at x=0 as follow:
If $x < 0$ than
\[\begin{gathered}
   \Rightarrow f(x) = {\sin ^{}}(|x|) + |x| \\
   \Rightarrow f(x) = {\sin ^{}}( - x) + ( - x) \\
\end{gathered} \]
\[ \Rightarrow f(x) = - {\sin ^{}}x + x\], as \[\sin ( - x) = - \sin x\]
Now, $Lf'(x) = - \cos x + 1$, as $\dfrac{{d(\sin (x))}}{{dx}} = \cos x$
$ \Rightarrow Lf'(0) = - 1 + 1 = 0$ , $\cos 0 = 1$
If $x > 0$ than
\[ \Rightarrow f(x) = {\sin ^{}}(|x|) + |x|\]
\[ \Rightarrow f(x) = {\sin ^{}}x + x\]
Now, $Rf'(x) = \cos x + 1$, as $\dfrac{{d(\sin (x))}}{{dx}} = \cos x$
$ \Rightarrow Rf'(0) = 1 + 1 = 2$ , $\cos 0 = 1$
$ \Rightarrow Lf'(0) \ne Rf'(0)$

Hence this function is not differentiable at x=0. Option C is incorrect
Now, check the differentiability of \[f(x) = {\sin ^{}}(|x|) - |x|\] at x=0 as follow:
If $x < 0$than
\[\begin{gathered}
   \Rightarrow f(x) = {\sin ^{}}(|x|) - |x| \\
   \Rightarrow f(x) = {\sin ^{}}( - x) - ( - x) \\
\end{gathered} \]
\[ \Rightarrow f(x) = - {\sin ^{}}x + x\], as \[\sin ( - x) = - \sin x\]
Now, $Lf'(x) = - \cos x + 1$, as $\dfrac{{d(\sin (x))}}{{dx}} = \cos x$
$ \Rightarrow Lf'(0) = - 1 + 1 = 0$ , $\cos 0 = 1$
If $x > 0$ than
\[ \Rightarrow f(x) = {\sin ^{}}(|x|) - |x|\]
\[ \Rightarrow f(x) = {\sin ^{}}x - x\]
Now, $Rf'(x) = \cos x - 1$, as $\dfrac{{d(\sin (x))}}{{dx}} = \cos x$
$ \Rightarrow Rf'(0) = 1 - 1 = 0$ , $\cos 0 = 1$
$ \Rightarrow Lf'(0) = Rf'(0)$
Hence this function is differentiable at x=0. Hence option D is correct.

So, D is the correct option.

Note: If a function is differentiable at a point then the limit of that function also exists. Moreover, the function is continuous at that point. While solving left-hand derivative and right-hand derivative make sure that you have taken first $x < 0$ for left-hand derivative and $x > 0$ for right-hand derivative.
While putting the value of x make sure first you find derivative of the function and after that put the value of x to find the value of left-hand as well as right-hand derivative.