<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] The function is polynomial everywhere except $x=\pm1,$ therefore it suffices to verify derivability with limits only at $x=1$ and $x=-1.$
At each of these values the limit exists (is equal to $0$), hence the function is differentiable on $\mathbb{R}.$
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