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derivative of tanφ=x/y

Guest Dec 5, 2014

Best Answer 

 #2
avatar+20681 
+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
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+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
avatar+20681 
+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
avatar+94140 
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
 #4
avatar+20681 
0

Hi Melody,

I have only guessed the question. Here is a possible answer.

heureka  Dec 5, 2014
 #5
avatar+94140 
0

Okay, thank you Heureka :)

Melody  Dec 5, 2014

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