<s>[INST] Proving a second derivative Given that $$y = \sin^3 x + \cos^3 x$$
prove that $$\frac{d^2 y}{dx^2} = \frac{3}{2} (\cos x + \sin x)(3 \sin 2x - 2)$$
I began with differentiating the equation as it is and it took me around twelve steps to reach the answer, and I'm guessing that factoring the sum of the cubes before differentiating wouldn't make it any easier. Is there a simpler approach to this problem?
Thanks in advance.
 [/INST] $y'=3\sin^2 x \cos x - 3\cos^2 x \sin x \\
= 3\sin x \cos x(\sin x - \cos x) \\
=\frac{3}{2}\sin(2x)(\sin x - \cos x)\\
y''=\frac{3}{2}\cos(2x)(2)(\sin x - \cos x)+\frac{3}{2}\sin(2x)(\cos x+\sin x)\\
=\frac{3}{2}(2(\cos^2 x-\sin^2 x)(\sin x - \cos x)+\sin(2x)(\cos x+\sin x))\\
=\frac{3}{2}(-2(\sin x - \cos x)(\sin x + \cos x)(\sin x - \cos x))+\sin(2x)(\sin x + \cos x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin x - \cos x)^2+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin^2 x -2\sin x \cos x + \cos^2 x)+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(1 -\sin(2x))+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(3\sin(2x)-2)\\$
As far as I can tell, the result given is wrong, so watch out :)
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