1. ## partial differentiation (trig)

Im trying to find the partial derivative of df/dy, but im not sure how to differentiate tan^-1 (x/y).
could someone check if i differentiated the log correctly?
Thanks!

2. Originally Posted by gomes

Im trying to find the partial derivative of df/dy, but im not sure how to differentiate tan^-1 (x/y).
could someone check if i differentiated the log correctly?
Thanks!

Let $p = tan^{-1} \frac{x}{y}$

$tanp = \frac{x}{y}$

Implicitly differentiate

$(sec^2 p) p^{ \prime} = - \frac{x}{y^2}$

$p^{ \prime} = - \frac{x}{y^2 sec^2 p }$

$p^{ \prime} = - \frac{x}{y^2 sec^2 (tan^{-1} \frac{x}{y} ) }$

3. thanx

4. That's correct but a very strange way of writing the answer!

I would, instead, have used the fact that the derivative of $tan^{-1}(x)$ is $\frac{1}{x^2+ 1}$. Let $u= \frac{x}{y}$ so that $\frac{\partial u}{\partial y}= \frac{\partial \frac{x}{y}}{\partial y}= -\frac{x}{y^2}$. Also $tan^{-1}(\frac{x}{y})= tan^{-1}(u)$ and $\frac{\partial tan^{-1}(u)}{\partial u}= \frac{1}{u^2+ 1}$.

By the chain rule, then, $\frac{\partial tan^{-1}(\frac{x}{y})}{\partial y}$ $= \frac{1}{u^2+ 1}\left(-\frac{x}{y^2}\right)=$ $-\frac{1}{\frac{x^2}{y^2}+ 1}\frac{x}{y^2}= -\frac{x}{x^2+ y^2}$.

But, as I said, AllenCuz's result is also correct- These answers are exactly the same.

Imagine a right triangle with angle $\theta$, "opposite side" of length x, and "near side" of length y. Then $tan(\theta)= \frac{x}{y}$ so [tex]/theta= tan^{-1}(\frac{x}{y}). But then the hypotenuse of the triangle has length $\sqrt{x^2+ y^2}$ and $sec(\theta)= sec(tan^{-1}(\frac{y}{x}))= \frac{\sqrt{x^2+ y^2}}{y}$ so that $sec^2(tan^{-1}(\frac{y}{x}))= \frac{x^2+ y^2}{y^2}$.

$-\frac{x}{y^2 sec(tan^{-2}(\frac{y}{x}))}= -\frac{x}{y^2}\frac{y^2}{x^2+ y^2}= -\frac{x}{x^2+ y^2}$.