Need help diffrenting trig problems In(x^2(3x-1))?

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Answer 1 · 2018-01-17T02:39:02

#d/dx[ln((x^2)(3x-1))]=(9x^2-2x)/(3x^3-x^2)#

Well there are no trig functions in this problem. Perhaps you meant to say logarithms? Anyways...

Apply the following technique:

#d/dx[ln(u)]=1/u*u'#

Given: #ln(x^2(3x-1))#

I would distribute the #x^2# to #3x-1# and rewrite this as

#ln(3x^3-x^2)#

So when differentiating, I'll let #u=3x^3-x^2# so now we'll find the deivative

#d/dx[ln(u)]=1/(3x^3-x^2)*color(red)(u')#

#color(red)(u'=9x^2-2x)#

So

#d/dx[ln(u)]=1/(3x^3-x^2)*9x^2-2x=(9x^2-2x)/(3x^3-x^2)#