How do you differentiate #(ln(x+1))^(cosx)#?

1 Answer
Sonnhard
Jun 23, 2018

#f'(x)=(ln(x+1))^cos(x)*(-sin(x)*ln(ln(x+1))+cos(x)/((x+1)*ln(x+1))#

Explanation:

To make thngs better we take the logaithm on both sides:
#ln(f(x))=cos(x)ln(ln(x+1))#
Now using the chain rule to differentiate both sides with respect to #x#:

#1/f(x)*f'(x)=-sin(x)ln(ln(x+1))+cos(x)*1/ln(x+1)*1/(x+1)#

multiplying by
#f(x)#

#f'(x)=(ln(x+1))^cos(x)*(-sin(x)*ln(ln(x+1))+cos(x)/((x+1)*ln(x+1)))#

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