Verifying the existence of a derivative of an absolute value function at x = 0.

Question

I've been struggling with this question for a while, and any advice would be greatly appreciated.

Answer 1

For $x>0$ we have $f(x)=x^4$, hence $f'(x)=4x^3$.

For $x<0$ we have $f(x)=-x^4$, hence $f'(x)=-4x^3$.

For $x=0$ we use the definition:

$$\frac{f(h)-f(0)}{h}=|h^3|=|h|^3 \to 0$$

as $h \to 0.$ Hence $f$ is differentiable at $0$ and $f'(0)=0.$

Answer 2

Your derivative calculation is correct but can be simplified in a different way for making the differentiability of $f$ much more obvious. By definition, $$\frac{d}{dx}(|x|)=\frac{x}{|x|}.$$ Now, if $f(x)=x|x^3|$, we can rewrite as $$f(x)=x|x^2 \cdot x|=x\cdot x^2\cdot |x|=x^3|x|.$$ Taking the derivative, we have \begin{align*} f'(x)&=3x^2 |x| + x^3\cdot\frac{x}{|x|}\\ &=\frac{3x^2\cdot|x|^2 + x^4}{|x|}\\ &=\frac{4x^4}{|x|}\\ &=\frac{4x^2\cdot|x|^2}{|x|}\\ &=4x^2 |x|. \end{align*} So, it is obvious now that the derivative exits at $x=0$. But as I said, one could derive at this result even from your calculation as follows: \begin{align*} \frac{4x^6}{|x^3|} &= \frac{4\cdot|x^3|^2}{|x^3|}\\ &= 4|x^3|\\ &= 4x^2|x|. \end{align*}