Q.If $f(x) = |x|^3$, show that $f''(x)$ exists for all real $x$ and find it.

Idea: write $|x|^3$ as a piecewise cubic, differentiate twice, and check $x=0$ from the limit definition.

Since $|x|=x$ for $x\ge 0$ and $|x|=-x$ for $x<0$,

First derivatives ($x\neq 0$): $f'(x)=3x^2$ for $x>0$ and $f'(x)=-3x^2$ for $x<0$; both give $f'(x)=3x|x|$. At $x=0$, $f'(0)=\lim_{h\to 0}\frac{|h|^3}{h}=\lim_{h\to 0}|h|\,h=0$. So $f'(x)=3x|x|$ everywhere. …

The plain absolute value $|x|$ has a corner at $x=0$ and is not differentiable there. But cubing it smooths that corner, so $|x|^3$ turns out to be twice differentiable everywhere. We show this by splitting into cases and checking $x=0$ carefully with the limit definition.

Step 1 — write $f$ piecewise

Since $|x|=x$ for $x\ge 0$ and $|x|=-x$ for $x<0$, and $(-x)^3=-x^3$,

Step 2 — first derivative for $x\neq 0$

Both cases are captured by $f'(x)=3x|x|$ (since $x|x|=x^2$ for $x>0$ and $-x^2$ for $x<0$).

Step 3 — check $f'(0)$

So $f'(x)=3x|x|$ holds for all $x$, including $0$.

Step 4 — second derivative for $x\neq 0$

Differentiate each piece:

$f''(x)=\begin{cases} 6x, & x>0\\ -6x, & x<0.\end{cases}$ …

Method: Differentiating an Absolute-Value Function Piecewise, and Checking the Seam

Whenever a function is built from $|x|$, split it into cases based on the sign of $x$, differentiate each case with the ordinary rules, and then check the transition point ($x=0$) separately using the limit definition of the derivative — never just by evaluating the piecewise formula there, since the two one-sided formulas might disagree.

Steps

Step 1: Rewrite the function piecewise using $|x|=x$ for $x\ge0$ and $|x|=-x$ for $x<0$

Step 2: Differentiate each piece using ordinary rules (for $x\neq0$)

Step 3: Combine the two pieces into a single formula if they match a common pattern (e.g. involving $|x|$ again)

Step 4: Check the point where the pieces meet using the limit definition

$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}.$ …

Common Mistakes

Mistake 1: Assuming $|x|^3$ inherits the non-differentiability of $|x|$ at $x=0$ without checking.

Why it's wrong: $|x|$ itself has a sharp corner at $0$ (left/right derivatives disagree), but cubing it smooths that corner out — the extra power changes the behaviour entirely, and this must be verified, not assumed by analogy. Correct approach: always run the limit-definition check at the seam point rather than pattern-matching to a simpler related function.

Mistake 2: Evaluating the piecewise formula at $x=0$ instead of using the limit definition. …