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What is the derivative of #f(x)=tan(1/lnx)#?

1 Answer

Dec 6, 2015

#=-(sec^2(1/lnx))/(xln^2x)#

Explanation:

According to the chain rule:

#f'(x)=sec^2(1/lnx)d/dx[1/lnx]#

#=sec^2(1/lnx)d/dx[ln^-1x]#

#=sec^2(1/lnx)xx-ln^-2x xx d/dx[lnx]#

#=sec^2(1/lnx)(1/-ln^2x)(1/x)#

#=-(sec^2(1/lnx))/(xln^2x)#

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