If #f(x)=ln(x^(cosx))#, then find #f'(2.1)#?

1 Answer
Feb 6, 2018

#f'(2.1)=-0.88085#

Explanation:

We should first find #f'(x)# using product rule, as the given functon can be written as a product of two functions,

as #f(x)=ln(x^cosx)=cosxlnx#

Therefore #f'(x)=-sinxlnx+cosx/x#

and #f'(2.1)=-sin2.1ln2.1+cos2.1/2.1#

= #-0.86321xx0.74194-0.50485/2.1#

= #-0.64045-0.24040#

= #-0.88085#