last one I promise... y=(ln(7x))/x^5.... what is y' and y''? I got y' correct. it is x^-6(1-5ln(7x))... cant figure out y''

Question

anonymous · Answer

try rewriting and using the chain.  ln(7x)x^(-5)

anonymous · Answer

first deriv.=  u'v+uv'
second deriv.= u''v+u'v'+u'v'+uv''

anonymous · Answer

u=ln(7x)
v=x^-5
v'=-5x^-6
v''=30x^-7

anonymous · Answer

i got that derivative correct. i just cant find y''

anonymous · Answer

Im blanking on ln deriv. for some reason im thinking its x^-1

anonymous · Answer

if you find all the parts just plug it in.

anonymous · Answer

u=ln(7x)
u'=7/7x
u"=quotient rule?

anonymous · Answer

yes ln deriv is 1/x or x^-1

anonymous · Answer

u'=7x^-1
u''=-7x^-2

anonymous · Answer

so second deriv.=
(-7x^(-2))(x^-5)+2(-5x^(-6))(7x^(-1))+ln(7x)30(x^(-7))

anonymous · Answer

clean it up and youre good

anonymous · Answer

I had just got that from your set up  of u"v+u'v'+u'v'+uv"! thanks! my professor never gave me that set up!

anonymous · Answer

welcome.