I've run into a problem while finding a directional derivative of a function. The function is as follows:
\begin{align} z &= \frac{x}{y} \\\\ \end{align} The task is to find its directional derivative at a point M whose coordinates are x=1, y=1 in the direction of a vector defined by f(x) that is perpendicular to the graph of f(x)=2x-1 at the point M. I found that the perpendicular in question is as follows: \begin{align} f(x) &= \frac{-x+3}{2} \\\\ \end{align}
Next, my job is to find cos(x) and cos(y) to plug into the formula:
\begin{align} z'_{l}=f'_{x}cos(x)+f'_{y}cos(y) \end{align}
The problem here is that my teacher and I have considered two different angles and, despite getting the same end results, do not agree about the intermediate steps. My cosines are the following: \begin{align} cos(x) &= -\frac{2}{\sqrt5} \\\\ cos(y) &= \frac{1}{\sqrt5} \end{align} I got cos(y) by considering that a vector defined by \begin{align} f(x) &= \frac{-x+3}{2} \\\\ \end{align} points from origin to the second quadrant, where angle x, the angle with the x-axis, is one that is greater than 90°, and angle y, the angle with the y-axis, is one that is less than 90°, it is adjacent to the y-axis as seen from the second quadrant. My teacher has instead considered both angles to belong to the first quadrant (they haven't considered the vector coming from the origin in the same way I have), which means both angles amount to over 90°. Their angle y is one that is adjacent to the y-axis as seen from the first quadrant. I was pretty sure my logic was very flawed while performing the computation, but I'd like to know what the truth here is for certain. The question is, which is the angle with the y-axis that is given by the function described above?
Thank you.