Answer
$y' = 2x \tan{\left(\dfrac{1}{x} \right)} - \sec^2{\left(\dfrac{1}{x}\right)}$
The zeros of $y'$ represent points on the graph of the original function $y$ where the tangent lines are horizontal.
Work Step by Step
The function is given by the equation $y=x^2 \tan{\left(\dfrac{1}{x}\right)}$
Using a computer algebra system, the derivative of the function is:
$y' = 2x \tan{\left(\dfrac{1}{x} \right)} - \sec^2{\left(\dfrac{1}{x}\right)}$
The red curve represents the graph of the function. The blue curve represents its derivative.
The zeros of $y'$ represent points on the graph of the original function $y$ where the tangent lines are horizontal.
