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

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

\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.