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

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Answer 1 · 2016-01-26T20:10:59

#f'(x) = -(cos((1)/(ln x)))/(xln^2x)#

We need to use the chain rule.

First, we can rename the function to make the calculations easier. So, we have a few functions called:

#f(u) = sin(u)#

where #u(t) = (1)/(t)#

and #t(x) = ln x#

So, the derivative of #f(x)# is (using the chain rule):

#f'(x) = (df)/(du)(du)/(dt)(dt)/(dx) #

We calculate these derivatives:

#(df)/(du) = cos(u)#

#(du)/(dt) = -(1)/(t^2)#

#(dt)/(dx) = (1)/(x)#

The final expression is the multiplication of them and the substitution of their previous definitions:

#f'(x) = cos(u)(-1)/(t^2)(1)/(x) = -cos((1)/(lnx))(1)/(xln^2x)#