To check differentiability at $x = 0$, we examine the left-hand and right-hand derivatives.
Step 1: Consider the function $f(x) = |x|$
Step 2: Evaluate the one-sided derivatives at $x = 0$
Conclusion:
Since $f'_-(0) \ne f'_+(0)$, the function $f(x) = |x|$ is not differentiable at $x = 0$.
Given: $f(x) = |x|$
Step 1: Piecewise definition of $f(x)$
$f(x) = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$
Step 2: Continuity check at $x = 0$
Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$
Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$
Function value: $f(0) = |0| = 0$
The function is continuous at $x = 0$.
Step 3: Differentiability check at $x = 0$
Left-hand derivative:
$f'_-(0) = \lim_{h \to 0^-} \dfrac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \dfrac{-h - 0}{h} = \lim_{h \to 0^-} (-1) = \mathbf{-1}$
Right-hand derivative:
$f'_+(0) = \lim_{h \to 0^+} \dfrac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \dfrac{h - 0}{h} = \lim_{h \to 0^+} (1) = \mathbf{1}$
Conclusion:
As $f'_-(0) e f'_+(0)$, the function $f(x) = |x|$ is not differentiable at $x = 0$.
If $e^y (x+1) = 1$, then find the value of $$ \frac{d^2 y}{dx^2} - \left(\frac{dy}{dx}\right)^2. $$