derivative of tanφ=x/y
$$\tan(\phi) = \frac{x}{y} \quad | \quad *\frac{y}{\tan(\phi)}\\ \\
y = \frac{x}{\tan(\phi)} \\ \\
y = \frac{1}{\tan(\phi)} * x \quad | \quad \frac{d()}{dx} \\ \\
y' = \frac{1}{\tan(\phi)} * \frac{d(x)}{dx} \quad | \quad \frac{d(x)}{dx} = 1\\ \\
y' = \frac{1}{\tan(\phi)} * 1\\ \\
y' = \frac{1}{ \tan(\phi) }\quad | \quad \tan(\phi) = \frac{x}{y} \\ \\
y' = \frac{1}{ \frac{x}{y} } \\ \\
y' = 1* \frac{y}{ x } \\ \\
\boxed{ y' = \frac{y}{ x } }$$
What is the exact function?
tanφ=x/y is an equation which you can solve for y, if x is the functions variable:
y (x) = x/(tanφ)
There derivative from this function is:
y ' (x) = 1/(tanφ)
-> But this is only the solution, if x is the input and y is the output of the function.
derivative of tanφ=x/y
$$\tan(\phi) = \frac{x}{y} \quad | \quad *\frac{y}{\tan(\phi)}\\ \\
y = \frac{x}{\tan(\phi)} \\ \\
y = \frac{1}{\tan(\phi)} * x \quad | \quad \frac{d()}{dx} \\ \\
y' = \frac{1}{\tan(\phi)} * \frac{d(x)}{dx} \quad | \quad \frac{d(x)}{dx} = 1\\ \\
y' = \frac{1}{\tan(\phi)} * 1\\ \\
y' = \frac{1}{ \tan(\phi) }\quad | \quad \tan(\phi) = \frac{x}{y} \\ \\
y' = \frac{1}{ \frac{x}{y} } \\ \\
y' = 1* \frac{y}{ x } \\ \\
\boxed{ y' = \frac{y}{ x } }$$
Thanks Heureka,
I will start by openly admiting that I do not understand this type of calculus.
BUT
When you were asked to find the derivative how did you know it was dy/dx that was wanted.
Also how come you have treated Φ like it is a constant. Why are you allowed to do that ?
Have you found a partial derivative?