We'll need to use the chain rule to find the derivative of tan(1/x)tan(1x).
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!
