We'll need to use the chain rule to find the derivative of #tan(1/x)#.
The chain rule is given by:
(#f@g(x))'=f'(g(x))g'(x)#
This may look a bit confusing, but in reality it's not so bad once you get the hang of it.
We can think of #tan(1/x)# as being composed of two functions, say #f(x)# and #g(x)#. We have #tan# as one function, and #1/x# as its own function inside of the first. We can think of #tan# as being #f(x)# and #1/x# as being #g(x)# in the above definition.
First, we take the derivative of the outermost function, #tan#, and leave the inner function as is. The derivative of the tangent is #sec^2#, so we have #sec^2(1/x)# as the first half of our derivative. This is #f'(g(x))# in the above definition. Next, we multiply by the derivative of the inside function, #1/x#.
I would rewrite #1/x# as #x^-1# and use the product rule to differentiate. We "bring down the power and reduce the power by one," giving #-x^-2#, which is equivalent to #-1/x^2#. This is #g'(x)# in the above definition.
Now we multiply these components together:
#d/dxtan(1/x)=-(sec^2(1/x))/x^2#
Hope that helps!
