Finding the third derivative

**calcbeg** · February 18th 2009, 06:53 PM

I really don't like trig functions so could someone give this the once over?

"Find f'''(x) of f(x) 6x^5 + 3xcosx"

f'(x) = 30 x^4 + 3x(-sinx) + 3cosx

f''(x) = 120x^3 + -3x(cosx) + -sinx

f'''(x) = 360x^2 + 3x(sinx) + -cosx

Did I do this right???

Thanks in advance

**Reckoner** · February 18th 2009, 07:43 PM

Originally Posted by calcbeg

I really don't like trig functions so could someone give this the once over?

"Find f'''(x) of f(x) 6x^5 + 3xcosx"

f'(x) = 30 x^4 + 3x(-sinx) + 3cosx

Looking good up to this point!

f''(x) = 120x^3 + -3x(cosx) + -sinx

I'm not sure where the last term came from, and you didn't handle the product rule correctly here.

$f'(x) = 30x^4-3x\sin x + 3\cos x$

So

$f''(x) = 120x^3-3(x\cos x + \sin x) - 3\sin x$

$= 120x^3-3x\cos x-3\sin x-3\sin x$

$= 120x^3-3x\cos x-6\sin x$

and

$f'''(x) = 360x^2-3(-x\sin x + \cos x)-6\cos x$

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