tan(x/y)=1 deifferentiation
Just to be clear, do you want to differentiate with respect to x?
\[\begin{split} \frac{d}{dx}\tan\left(\frac{x}{y}\right) &= \frac{\frac{d}{dx}\left(\frac{x}{y}\right)}{\cos^2\left(\frac{x}{y}\right)} = \frac{\frac{1\cdot y-x\cdot y'}{y^2}}{\cos^2\left(\frac{x}{y}\right)} = \frac{y-x\cdot y'}{y^2\cos^2\left(\frac{x}{y}\right)} \end{split}\]
I'm using the formula for the derivative of tan(x) and the chain rule in the first step. In the second step, I'm using the quotient rule, and in the third step I'm just simplifying.
Here I define it as Θ=x/y \[\tan \theta = 1\] \[\theta = \frac{ \pi }{ 4 } + n \pi (n:integer)\] So this question is \[\frac{ x }{ y } = \frac{ \pi }{ 4 } + n \pi \] \[y = \frac{ 4 }{ (4+n) \pi } x\] \[\frac{ dy }{ dx } = \frac{ 4 }{ (4+n) \pi }\]
I missed `= 0' in the end. Good idea Tsuyo.
You made a mistake when going to \(y = \frac{ 4 }{ (4+n) \pi } x\). You should have written \(y = \frac{ 4 }{ (1+4n) \pi } x\).
Thank you, JoelSjogren! The correct answer is \[\frac{ dy }{ dx } = \frac{ 4 }{ (1+4n) \pi }\]
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