# Thread: High order derivative problem

1. ## High order derivative problem

I having trouble figuring this problem out:

Find $f^\prime^\prime(2)$ for $tsin\frac{\pi}{t}$

So using the product rule, $f^\prime(t)= tcos\frac{\pi}{t}+sin\frac{\pi}{t}$

Using the sum rule, $f^\prime^\prime(t)=t(-sin\frac{\pi}{t})+cos\frac{\pi}{t}$

so plugging in $f^\prime^\prime(2)=2(-(1))+0=-2$

Does that seem right?

Any help is greatly appreciated!

2. Originally Posted by ascendancy523
I having trouble figuring this problem out:

Find $f^\prime^\prime(2)$ for $tsin\frac{\pi}{t}$

So using the product rule, $f^\prime(t)= tcos\frac{\pi}{t}+sin\frac{\pi}{t}$

Using the sum rule, $f^\prime^\prime(t)=t(-sin\frac{\pi}{t})+cos\frac{\pi}{t}$

so plugging in $f^\prime^\prime(2)=2(-(1))+0=-2$

Does that seem right?

Any help is greatly appreciated!
You need to apply the chain rule.

$f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{d}{dt}\lef t(\frac{\pi}{t}\right) + \sin\frac{\pi}{t}$

etc.

Also, IF the first derivative were what you wrote, you (still) would have gotten the wrong second derivative; the first term would require the product and chain rules, and the second term would require the chain rule.

3. Originally Posted by undefined
You need to apply the chain rule.

$f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{d}{dt}\lef t(\frac{\pi}{t}\right) + \sin\frac{\pi}{t}$

etc.

Also, IF the first derivative were what you wrote, you (still) would have gotten the wrong second derivative; the first term would require the product and chain rules, and the second term would require the chain rule.
So $f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{-\pi}{t^2}+sin\frac{\pi}{t}$?

4. Originally Posted by ascendancy523
So $f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{-\pi}{t^2}+sin\frac{\pi}{t}$?
Yep.