Find \[\dfrac{d}{{dx}}\tan \left( {\dfrac{1}{x}} \right)\]

Hint: We are given with a function but in a composition form. In this, a function involves one more function. So we will use chain rule to find the derivative of it. We will use the derivative of tan function and derivative of power function.
Formula used:
Chain rule : \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)\]
\[\dfrac{d}{{dx}}\tan \left( x \right) = {\sec ^2}x\]
\[\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = \dfrac{{ - 1}}{{{x^2}}}\]

Complete step-by-step answer:
Given is \[\dfrac{d}{{dx}}\tan \left( {\dfrac{1}{x}} \right)\]
Now here two functions are in composition.
So we will use the chain rule given above to find the derivative. First we will find the derivative of tan but the function will be the composition function. So,
\[\dfrac{d}{{dx}}\tan \left( {\dfrac{1}{x}} \right) = \dfrac{d}{{dx}}\tan \left( {\dfrac{1}{x}} \right) \times \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\]
We know that, \[\dfrac{d}{{dx}}\tan \left( x \right) = {\sec ^2}x\]
\[ = {\sec ^2}\left( {\dfrac{1}{x}} \right) \times \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\]
Now the derivative of second function, \[\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = \dfrac{{ - 1}}{{{x^2}}}\]
\[ = {\sec ^2}\left( {\dfrac{1}{x}} \right) \times \left( {\dfrac{{ - 1}}{{{x^2}}}} \right)\]
Now we will rearrange the terms as,
\[ = \dfrac{{ - {{\sec }^2}\left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\]
Thus this is the answer.
So, the correct answer is “\[ \dfrac{{ - {{\sec }^2}\left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\]”.

Note: Note that the derivative of sum and difference of two functions results in the sum and difference of their derivatives only. But this is not in case of product and division. That’s why product rule, chain rule and quotient rule are used. Also note that the chain rule is like two embedded functions and product rule has two separate functions.