<s>[INST] Show that $f(x) = |1-x^2|^3$ is a differentiable function. $f: \mathbb{R} \rightarrow \mathbb{R}$
$x \rightarrow |1-x^2|^3$
Show that f is a differentiable function and calculate its derivative.
Check whether f if is a continuous function.

$f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0} = lim_{x\rightarrow x_0}\frac{|1-x^2|^3-|1-x_0^2|^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^2}^3- \sqrt{(1-x_0^2)^2}^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^6}- \sqrt{(1-x_0^2)^6}}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0}$
But I didn't get any further. How to get the $x-x_0$ out? I tried multiplying out the numerator and then using polynomial division but I didn't get a solution.
So I tried calculating the derivative directly, but that doesn't show it generally.
$f(x) = |1-x^2|^3 = (\sqrt{(1-x^2)^2})^3$
$f'(x) \\= 3 \cdot (\sqrt{(1-x^2)^2})^2 \cdot \frac{1}{2} \cdot ((1-x^2)^2)^{-\frac{1}{2}} \cdot 4x\cdot (1-x^2) \\ = 3 \cdot |1-x^2|^2 \cdot \frac{1}{|1-x^2|} \cdot 2x \cdot (1-x^2) \\= |1-x^2| \cdot 6x \cdot (1-x^2)$
I looked at the graph. The graph looks like a "W" but with "soft turns". If I can prove that the function is differentiable in every point then it is a continuous function.
So how do I prove it using this definition: $f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0}$ ?
 [/INST] Two cases: $|x_0|=1$ and $|x_0|\ne 1$ - that's where $|1-x^2|$ does not have a derivative.
For the first one (let us take $x_0=1$; the case $x_0=-1$ is done likewise) you get by definition
$$\lim_{x\to 1}\frac{|1-x^2|^3}{x-1} = \lim_{h\to 0}\frac{|1-(1+h)^2|^3}{h} = \lim_{h\to 0}\frac{|h^2+2h|^3}{h} = \lim_{h\to 0}\frac{|h|^3|h+2|^3}{h} = \lim_{h\to 0}\frac{|h|^3|h+2|^3}{h} =  \lim_{h\to 0} \left(sgn (h)\cdot h^2|h+2|^3\right) = 0.$$
In the case where $x_0\ne \pm 1$ you can  say that for $x$ sufficiently close to $x_0$ you can expand absolute value; for example, for $|x|<1$ you get $|1-x^2| = 1 - x^2$. After this transformation proceed as you would when proving that polynomials are differentiable. 
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