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

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Answer 1 · 2016-06-26T20:06:55

#= - sec^2 (1/x) * (1/x^2)#

we use the chain rule

to make it clear we can say let #w = 1/x#

so #y = tan w# means that #dy/(dw) = sec^2 w#

and #(dw)/dx = d/dx (1/x) = - 1/x^2#

the chain rule tells us that

#dy/dx = dy/(dw) * (dw)/dx#

#= sec^2 w * (- 1/x^2)#

#= sec^2 (1/x) * (- 1/x^2)#

#= - sec^2 (1/x) * (1/x^2)#