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# derivative of tanφ=x/y

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derivative of tanφ=x/y

Guest Dec 5, 2014

#2
+19054
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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
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#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
+19054
+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
#3
+91900
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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
+19054
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Hi Melody,

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

heureka  Dec 5, 2014
#5
+91900
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Okay, thank you Heureka :)

Melody  Dec 5, 2014

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