# High order derivative problem

• Jun 22nd 2010, 07:46 AM
ascendancy523
High order derivative problem
I having trouble figuring this problem out:

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

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

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

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

Does that seem right?

Any help is greatly appreciated!
• Jun 22nd 2010, 08:06 AM
undefined
Quote:

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

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

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

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

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

Does that seem right?

Any help is greatly appreciated!

You need to apply the chain rule.

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

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.
• Jun 22nd 2010, 11:36 AM
ascendancy523
Quote:

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

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

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 $\displaystyle f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{-\pi}{t^2}+sin\frac{\pi}{t}$?
• Jun 22nd 2010, 11:48 AM
undefined
Quote:

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

Yep.