Values of k such that $f(x)=x^k|x|$ is 3 times differentiable at the origin We must find values of k such that  $f(x)=x^k|x|$ is 3 times differentiable at the origin.
Firstly we can view the question as a product of two functions $g(x)=x^k$ and $h(x)=|x|$
You may obtain the result of triple product rule as a function in terms of $f'$s and $h'$s
It is clear to me that $h'$ nor $h''$ is not differentiable at the origin since the denominator of both contains a x, however I was unable to obtain a third derivative of h
Thus all terms containing a h prime terms must be cancelled out by the f components, which clearly have zeros at $k=0,1,2$ respectively of their times differentiated
This leads to the only solution being $k=0$ which cancels out all of the absolute value terms,
but I suspect that this cannot be correct, is my intuition correct or is there some piece I am missing to solve this?
EDIT : should be product of functions not composition
 A: $f(x)$ is differentiable at the origin if $\lim_{x\to 0}\frac{f(x)-f(0)}{x}$ exists.
Here we have $$\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}x^k\frac{|x|}{x}=\lim_{x\to 0}x^k\operatorname{sign}(x),$$
where $\operatorname{sign}(x)$ is $+1$ if $x>0$ and $-1$ for $x<0$.
If $k=0$ then this limit clearly does not exist, but if $k>0$ then the limit is $0$ because one of the factors tends to $0$ and the other is bounded.
Thus for $k>0$, $f(x)$ is differentiable. What is its derivative? Well, we know $f'(0)=0$. Since $f(x)=x^{k+1}$ for $x>0$ the derivative in the positive region is $(k+1)x^k$, and since $f(x)=-x^{k+1}$ for $x<0$ the derivative for $x$ negative is $-(k+1)x^k$. Putting all these together we get
$$f'(x)=(k+1)x^{k-1}|x|.$$
Now continue in this manner to determine when $f(x)$ is twice, three times, etc., differentiable.
A: First write $f$ piecewise so we can calculate the second derivative:
$$f(x) = \begin{cases} x^{k + 1} & x > 0 \\ -x^{k + 1} & x < 0 \end{cases}$$
Then
$$f''(x) = \begin{cases} (k^2 + k)x^{k - 1} & x > 0 \\ -(k^2 + k)x^{k - 1} & x < 0 \end{cases}$$
Then you need
$$ f'''(0) = \lim_{x \to 0} \frac{f''(x)}{x} $$
to exist.
Notice we do not calculate $f'''(x)$ in the same way we calculated $f''(x)$ since we only care about $f'''(0)$ and not for $f''$ to be continuously differentiable.
By the way: it's a bit hidden but why do we know that $f''(0)$ must be $0$ if it exists? Or if $f'''(0)$ exists?
