<s>[INST] Partial derivative of $f(x,y) = (x/y) \cos (1/y)$ So I'm not really sure whether I'm correct as several people are saying some of my syntax is wrong, where others are saying I have a wrong answer. I have checked my answer using Wolfram Alpha and it appears to be correct; could anyone please confirm/clarify?
Calculate the partial Derivative $\frac{\partial f}{\partial y}$ of
$$f(x,y) = \frac{x}{y} \cos\left(\frac{1}{y}\right).$$
Is this correct? (Slightly reduced working because it's super long to type out)
$x$ is constant
$$\frac{\partial f}{\partial y} \left(\frac{\cos(\frac{1}{y})}{y}\right)$$
Quotient Rule
$$\frac{\dfrac{\partial f}{\partial y} \left(\dfrac{\cos(\frac{1}{y})}{y}\right)y-\dfrac{\partial f}{\partial y}(y) \cos(\frac{1}{y})}{y^2}$$
Chain Rule
$$\frac{\partial f}{\partial y} \left(\cos(\tfrac{1}{y})\right) = \frac{\sin(\frac{1}{y})}{y^2}$$
$$x\frac{\frac{\sin\frac{1}{y}}{y^2}y-1\cos(\frac{1}{y})}{y^2}$$
Simplified Answer
$$\frac{x \left(\sin\frac{1}{y} -y\cos(\frac{1}{y})\right)}{y^3}$$
 [/INST] $$\frac { \partial f\left( x,y \right)  }{ \partial y } =\frac { \partial \left( \frac { x }{ y } \cos { \frac { 1 }{ y }  }  \right)  }{ \partial y } =x\frac { \partial \left( \frac { 1 }{ y }  \right)  }{ \partial y } \cos { \frac { 1 }{ y } +\frac { x }{ y } \frac { \partial \left( \cos { \frac { 1 }{ y }  }  \right)  }{ \partial y }  } =\\ =-\frac { x }{ { y }^{ 2 } } \cos { \frac { 1 }{ y } +\frac { x }{ y^{ 3 } } \sin { \frac { 1 }{ y }  }  } =\frac { -x\left( y\cos { \frac { 1 }{ y } -\sin { \frac { 1 }{ y }  }  }  \right)  }{ { y }^{ 3 } } $$
</s>