#2**+5 **

**derivative of tanφ=x/y**

**$$\tan(\phi) = \frac{x}{y} \quad | \quad *\frac{y}{\tan(\phi)}\\ \\ y = \frac{x}{\tan(\phi)} \\ \\ y = \frac{1}{\tan(\phi)} * x \quad | \quad \frac{d()}{dx} \\ \\ y' = \frac{1}{\tan(\phi)} * \frac{d(x)}{dx} \quad | \quad \frac{d(x)}{dx} = 1\\ \\ y' = \frac{1}{\tan(\phi)} * 1\\ \\ y' = \frac{1}{ \tan(\phi) }\quad | \quad \tan(\phi) = \frac{x}{y} \\ \\ y' = \frac{1}{ \frac{x}{y} } \\ \\ y' = 1* \frac{y}{ x } \\ \\ \boxed{ y' = \frac{y}{ x } }$$**

heureka
Dec 5, 2014

#1**+5 **

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.

Guest Dec 5, 2014

#2**+5 **

Best Answer

**derivative of tanφ=x/y**

**$$\tan(\phi) = \frac{x}{y} \quad | \quad *\frac{y}{\tan(\phi)}\\ \\ y = \frac{x}{\tan(\phi)} \\ \\ y = \frac{1}{\tan(\phi)} * x \quad | \quad \frac{d()}{dx} \\ \\ y' = \frac{1}{\tan(\phi)} * \frac{d(x)}{dx} \quad | \quad \frac{d(x)}{dx} = 1\\ \\ y' = \frac{1}{\tan(\phi)} * 1\\ \\ y' = \frac{1}{ \tan(\phi) }\quad | \quad \tan(\phi) = \frac{x}{y} \\ \\ y' = \frac{1}{ \frac{x}{y} } \\ \\ y' = 1* \frac{y}{ x } \\ \\ \boxed{ y' = \frac{y}{ x } }$$**

heureka
Dec 5, 2014

#3**0 **

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?

Melody
Dec 5, 2014