Find the derivative with respect to $x$ of $$\cos^{-1}\left(\frac{\sqrt{1 - x^3} - \sqrt{1 + x^3}}{2}\right).$$
Here's my work:
Substituting $x^3 = \cos(2\theta):$ $$\begin{aligned}\cos^{-1}\left(\frac{\sqrt{1 - x^3} - \sqrt{1 + x^3}}{2}\right) &= \cos^{-1}\left(\frac{\sqrt{1 - \cos(2\theta)} - \sqrt{1 + \cos(2\theta)}}{2}\right)\\& = \cos^{-1}\left(\frac{\sqrt{2\sin^2\theta} - \sqrt{2\cos^2\theta}}{2}\right)\\& = \cos^{-1}\left(\frac{1}{\sqrt{2}}\sin\theta - \frac{1}{\sqrt2}\cos\theta\right) \\& = \cos^{-1}\left(\sin\theta\cos(\pi/4) - \sin(\pi/4)\cos\theta\right) \\& = \cos^{-1}\sin\left(\theta - \frac{\pi}{4} \right)\\& = \cos^{-1}\cos\left(\frac{\pi}2 - \theta + \frac{\pi}{4} \right)\\& = \frac{3\pi}4 - \theta \\& = \frac{3\pi}4 - \frac12\cos^{-1}\left(x^3\right).\end{aligned}$$
Differentiating this gives $$\boxed{\frac{3x^2}{2\sqrt{1 - x^6}}}.$$ So far so good... My textbook also shows this answer.
But the problem is that this, this, this, and the symPy package of Python all give this clunky result $$\boxed{-\frac{-\frac{3x^{2}}{2\sqrt{1-x^{3}}}-\frac{3x^{2}}{2\sqrt{1+x^{3}}}}{2\sqrt{1-\frac{1}{4}\left(\sqrt{1-x^{3}}-\sqrt{1+x^{3}}\right)^{2}}}}$$ (not even neutralising the three negative signs) instead of the above simpler one.
WolframAlpha and Desmos shows that these two expressions have the same domain $(-1,1)$ and are identically equal.
Why do these software not give the answer in simplified form?