How do you differentiate #f(x)=cot(ln2x) # using the chain rule?

1 Answer
Bio
Feb 27, 2016

#f'(x) = -csc^2(ln(2x))/x#

Explanation:

Substitute #u = ln(2x) = ln(2) + ln(x)#.

#frac{"d"u}{"d"x} = 1/x#

Now substitute #x# and differentiate.

#f'(x) = frac{"d"}{"d"x}(cot(ln(2x)))#

#= frac{"d"}{"d"x}(cot(u))#

#= frac{"d"}{"d"u}(cot(u)) frac{"d"u}{"d"x}#

#= -csc^2(u) 1/x#

#= -csc^2(ln(2x))/x#

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