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I want to find ${\theta}'(t)$ so by the chain rule I want $${\theta}'(x)*x'(t)+{\theta}'(y)*y'(t)$$. I know $${\theta}=arctan(y/x)$$ but how do I partially differentiate theta w.r.t x and y?

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- Thread starter
- Banned
- #1

I want to find ${\theta}'(t)$ so by the chain rule I want $${\theta}'(x)*x'(t)+{\theta}'(y)*y'(t)$$. I know $${\theta}=arctan(y/x)$$ but how do I partially differentiate theta w.r.t x and y?

- Feb 13, 2012

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If $\displaystyle \theta(x,y)= \tan^{-1} \frac {y}{x}$ then is...

I want to find ${\theta}'(t)$ so by the chain rule I want $${\theta}'(x)*x'(t)+{\theta}'(y)*y'(t)$$. I know $${\theta}=arctan(y/x)$$ but how do I partially differentiate theta w.r.t x and y?

$\displaystyle \theta^{\ '}_{x}= \frac{- \frac{y}{x^{2}}}{1+ (\frac{y}{x})^{2}}= - \frac{y}{x^{2}+y^{2}}$

$\displaystyle \theta^{\ '}_{y}= \frac{\frac{1}{x}}{1+ (\frac{y}{x})^{2}}= \frac{x}{x^{2}+y^{2}}$

Kind regards

$\chi$ $\sigma$