# Calculus: Second Order Derivative with Sin and Cos

$D^2 [\sin(\theta)+\cos^3(\theta)]$

The answer should be $-\sin(\theta)+6\sin^2 (\theta)\cos(\theta)-3\cos^3 (\theta)$

I understand it's applying $D$ twice, but I can't tell which rules to use. Since it's addition I can do $D[\sin(\theta)]+D[\cos^3 (\theta)]$ $= [\cos(\theta)]+3[-\sin^{3-1}(\theta)]$ $= \cos(\theta)-3\sin^2(\theta)$ Then $D^2=-\sin(\theta)-6\sin(\theta)$

What the hell am I missing? Should I apply the chain rule somewhere in there?

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$\sin(\theta)$ is a function, and $x^3$ is a function. Hence, in order to find the derivative of $\sin^3(\theta)$ you have to apply the chain rule. That's where your mistake is. (When you write $(x^3)'=3x^2$ you actually should write $(x^3)'=3x^2\cdot(x)'$, but $(x)'=1$. Here it's not true)

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You are evalutating $$\frac{d^2}{d\theta}(\sin \theta + \cos^3 \theta)$$

I suggest it helps when you are confused to write out the terms fully. So for the second term you use the chain rule (which you are doing already)

Write $y=\cos^3 \theta$ (Looking only at the second term). We know that $$\frac{dy}{d \theta} = \frac{dy}{du}\frac{du}{d \theta}$$ where $u = \cos \theta$. But then it is clear that $$\frac{du}{d \theta} = -\sin \theta$$ and $$\frac{dy}{du} = 3u^2 = 3 \cos^2 \theta$$.

Hence $$\frac{d}{d\theta}(\sin \theta + \cos^3 \theta)=\cos \theta - 3 \cos^2 \theta \sin \theta$$

You must then repeat this process, using the product and chain rule. Again if you write out the terms fully, I think it will help a lot!

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so then it would be -sinθ-3(cos^2θ)(cosθ)+sinθ(-6cosθ)(-sinθ) =-sinθ+6cosθsin2θ-3cos3θ Ah thank you! – campbell May 3 '11 at 7:16