Question

How do you find the derivative of `tan(1/x)`?

Answer 1

`-(sec^2(1/x))/x^2=-1/x^2sec^2(1/x)`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larr" chain rule"`

`rArrd/dx(tan(1/x))`

`=sec^2(1/x)xxd/dx(x^-1)`

`=sec^2(1/x)xx-x^-2`

`=-(sec^2(1/x))/x^2=-1/x^2sec^2(1/x)`