How do you differentiate #f(x)=(lnx)(ln3x) # using the product rule?

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Answer 1 · 2015-12-23T19:28:33

#f'(x)=(ln3x+lnx)/x#

According to the product rule:

#f'(x)=ln3xd/dx[lnx]+lnxd/dx[ln3x]#

Find each derivative individually:

#d/dx[lnx]=1/x#

Chain rule coming up:

#d/dx[ln3x]=1/(3x)*d/dx[3x]=3/(3x)=1/x#

(Another way to find this is to separate #ln3x# into #ln3+lnx#. Since #ln3# is a constant, its derivative is #0# leaving just #d/dx[lnx]=1/x#.)

Plug these back in.

#f'(x)=ln3x(1/x)+lnx(1/x)#

#f'(x)=(ln3x+lnx)/x#