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.
 A: Note that $$y=\sin^3 x+ \cos^3 x=(\sin x+\cos x)(\sin^2 x - \sin x\cos x+ \cos^2 x)=(\sin x+\cos x)(1- \frac12\sin 2x)$$
Then $$y'=-(\sin x+\cos x)\cos 2x+(\cos x-\sin x)(1- \frac12\sin 2x)$$ and $$y''=2(\sin x+\cos x)\sin 2x-2(\cos x-\sin x)\cos 2x-(\sin x+\cos x)(1- \frac12\sin 2x)$$
For the middle term $\cos 2x= \cos^2 x-\sin^2 x=(\cos x +\sin x)(\cos x-\sin x)$ and $(\cos x- \sin x)^2=\cos^2 x-2\cos x\sin x+ \sin^2 x=1-\sin 2x$
This gets you a factor $\cos x+ \sin x$ and the remaining factor in terms of $\sin 2x$, so it remains to simplify.
A: $y'=3\sin^2 x \cos x - 3\cos^2 x \sin x \\
= 3\sin x \cos x(\sin x - \cos x) \\
=\frac{3}{2}\sin(2x)(\sin x - \cos x)\\
y''=\frac{3}{2}\cos(2x)(2)(\sin x - \cos x)+\frac{3}{2}\sin(2x)(\cos x+\sin x)\\
=\frac{3}{2}(2(\cos^2 x-\sin^2 x)(\sin x - \cos x)+\sin(2x)(\cos x+\sin x))\\
=\frac{3}{2}(-2(\sin x - \cos x)(\sin x + \cos x)(\sin x - \cos x))+\sin(2x)(\sin x + \cos x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin x - \cos x)^2+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin^2 x -2\sin x \cos x + \cos^2 x)+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(1 -\sin(2x))+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(3\sin(2x)-2)\\$
As far as I can tell, the result given is wrong, so watch out :)
A: Your first step is to use the chain rule to take the derivative:
$$y' = 3 \cos x \sin^2 x - 3 \sin x \cos^2 x$$
You can avoid the product rule in the next step by changing $\sin^2 x$ to $(1-\cos^2 x)$ and $\cos^2 x$ to $(1-\sin^2 x)$. This gives an expression in four terms that needs no product rule.
Is this simpler enough for you?
