Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$y'=3\sin^2 x \cos x - 3\cos^2 x \sin x \\= 3\sin x \cos x(\sin^2 x - cos^2 x) \\= – Paul Sundheim Sep 1 '14 at 13:24
up vote 3 down vote accepted

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.

share|cite|improve this answer
That was a typo from me, sorry. – ZaIROuS Sep 1 '14 at 14:15
@ZaIROuS I've deleted the comment now you have corrected the question. – Mark Bennet Sep 1 '14 at 14:18

$y'=3\sin^2 x \cos x - 3\cos^2 x \sin x \\ = 3\sin x \cos x(\sin x - \cos x) \\ =\frac{3}{2}\sin(2x)(\sin x - \cos x)\\ y''=\frac{3}{2}\cos(2x)(2)(\sin x - \cos x)+\frac{3}{2}\sin(2x)(\cos x+\sin x)\\ =\frac{3}{2}(2(\cos^2 x-\sin^2 x)(\sin x - \cos x)+\sin(2x)(\cos x+\sin x))\\ =\frac{3}{2}(-2(\sin x - \cos x)(\sin x + \cos x)(\sin x - \cos x))+\sin(2x)(\sin x + \cos x))\\ =\frac{3}{2}(\sin x + \cos x)(-2(\sin x - \cos x)^2+\sin(2x))\\ =\frac{3}{2}(\sin x + \cos x)(-2(\sin^2 x -2\sin x \cos x + \cos^2 x)+\sin(2x))\\ =\frac{3}{2}(\sin x + \cos x)(-2(1 -\sin(2x))+\sin(2x))\\ =\frac{3}{2}(\sin x + \cos x)(3\sin(2x)-2)\\$

As far as I can tell, the result given is wrong, so watch out :)

share|cite|improve this answer
You're right, that was a typo. – ZaIROuS Sep 1 '14 at 14:16

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?

share|cite|improve this answer
Shouldn't I factor it to obtain $\sin 2x$? – ZaIROuS Sep 1 '14 at 13:50
$3\sin x\cos x (\sin x - \cos x) = \frac{3}{2} \sin 2x (\sin x - \cos x )$ – ZaIROuS Sep 1 '14 at 13:52
@ZaIROuS: Factoring out $\sin x \cos x$ out of the expression for $y'$ then changing that to $\frac 12 \sin 2x$ would also work but does not avoid the product rule. I suppose it is a matter of taste which is best: the product rule from the original expression, my suggestion, or the $\sin 2x$ route. – Rory Daulton Sep 1 '14 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.