How do you differentiate #F(t)=ln[(2t+1)^3/(3t-1)^4]#?

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Answer 1 · 2016-08-08T14:33:33

#=- (6(t+ 3))/((2t+1)(3t-1))#

#F(t)=ln[(2t+1)^3/(3t-1)^4]#

break it up to make it easier....
#=ln(2t+1)^3 - ln (3t-1)^4#

#=3ln(2t+1) - 4ln (3t-1)#

#d/dt (3ln(2t+1) - 4ln (3t-1))#

#=6/(2t+1) - 12/(3t-1)#

#=(18t - 6 - 24t -12)/((2t+1)(3t-1))#

#=(-6t - 18)/((2t+1)(3t-1))#

#=- (6(t+ 3))/((2t+1)(3t-1))#