High order derivative problem

**ascendancy523** · June 22nd 2010, 07:46 AM

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!

**undefined** · June 22nd 2010, 08:06 AM

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.

**ascendancy523** · June 22nd 2010, 11:36 AM

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 $f'(t)=t\cdot\cos\frac{\pi}{t}\cdot\frac{-\pi}{t^2}+sin\frac{\pi}{t}$ ?

**undefined** · June 22nd 2010, 11:48 AM

Originally Posted by ascendancy523

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

Yep.

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