# Thread: Need some help finding the 2nd deriv of this problem.

1. ## Need some help finding the 2nd deriv of this problem.

F(theta)= (cos(theta pi))^ -2 find 2nd derivative of F(2). I have tried it different ways. I have worked it from the original to the 1st then 2nd derivative of theta. At the end I plugged 2 in and that was no good. I tried taking derivative of 2 from the 1st to the second and that hasnt worked either. Can some please explain where I am wrong. If i replace theta with 2 then would it not be (cos(2pi))^ -2 and wouldn't that be the same as 1/((cos(2 pi))^2 and then use the quotient rule for the 1st derivative and go from there? Who knows my brain is fried and help would be awesome.

2. ## Re: Need some help finding the 2nd deriv of this problem.

We are given:

$\displaystyle F(\theta)=(\cos(\pi\theta))^{-2}$

and we are asked to find:

$\displaystyle F''(2)$

To find the first derivative, we may use the power and chain rules:

$\displaystyle F'(\theta)=-2(\cos(\pi\theta))^{-3}(-\pi\sin(\pi\theta))=2\pi\sin(\pi\theta)(\cos(\pi \theta))^{-3}$

To find the second derivative, we may use the product, power and chain rules:

$\displaystyle F''(\theta)=2\pi\left(\sin(\pi\theta)\left(-3(\cos(\pi\theta))^{-4}(-\pi\sin(\pi\theta)) \right)+\pi\\cos(\pi\theta)(\cos(\pi\theta))^{-3} \right)$

$\displaystyle F''(\theta)=2\pi^2(\cos(\pi\theta))^{-4}\left(3\sin^2(\pi\theta)+\cos^2(\pi\theta) \right)$

$\displaystyle F''(\theta)=2\pi^2\sec^4(\pi\theta)\left(2-\cos(2\pi\theta) \right)$

Hence:

$\displaystyle F''(2)=2\pi^2$

3. ## Re: Need some help finding the 2nd deriv of this problem.

Thank you very much. Im still a little bit confused how did you factor out the pi in the first deriv. -pi sin (theta pi)

4. ## Re: Need some help finding the 2nd deriv of this problem.

It stems from the application of the chain rule.

$\displaystyle \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}\frac{d}{d\theta}\left(\cos(\pi\theta) \right)$

$\displaystyle \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}(-\sin(\pi\theta))\frac{d}{d\theta}(\pi\theta)$

$\displaystyle \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=-2(cos(\pi\theta)^{-3}(-\sin(\pi\theta))(\pi)$

$\displaystyle \frac{d}{d\theta}\left((cos(\pi\theta)^{-2} \right)=2\pi\sin(\pi\theta)(cos(\pi\theta)^{-3}$