Help with Partial Derivatives and Chain rule

Question

The question is Determine $f_{xy}$ when $f = ytan^{-1}(xy)$ I know that there is chain rule somewhere in here, but I don't understand where it comes from.

Answer 1

To find $f_{xy}$ means we first have to find $\frac{\partial f}{\partial x}=g$, say, and then we do $\frac{\partial g}{\partial y}$.

Now $\frac{\partial f}{\partial x}= \frac{d}{dx}(y tan^{-1}(xy)$.

This means we can treat the $y$ as a constant, and use the chain rule, with $v=xy$, and $u=ytan^{-1}(v)$.

Depending on your knowledge of derivatives of given functions, this could be easy or difficult. I will state the rule that if $y=tan^{-1}(x)$, then $\frac{dy}{dx}=\frac{1}{1+x^2}$. I assume you have seen this if you are computing partial derivatives, but if not, here is a link to show the simple proof: http://www.themathpage.com/aCalc/inverse-trig.htm#arctan

The chain rule states that $\frac{d}{dx}(u(v(x))=u'(v(x))\:\text{x}\:v'(x)$. In our case, we have $$\frac{du}{dx}=\frac{du}{dv}\:\text{x}\:\frac{dv}{dx}=\frac{y}{1+v^2}\:\text{x}\:y$$ $$=\frac{y^2}{1+(xy)^2}=g$$

The next step is to differentiate $g$ with respect to $y$, in which case we need the quotient rule.

Answer 2

$f_{xy}= \frac{\partial\frac{\partial f}{\partial x}}{\partial y} $

Here: $ \frac{\partial f}{\partial x}= \frac{y^2}{1+x^2y^2}$ So, $f_{xy}(x,y)=\frac{2y(1+x^2y^2)-2yx^2(y^2)}{(1+x^2y^2)^2}= \frac{2y}{(1+x^2y^2)^2}$

P.S Please use $artcan(x)$ instead of $tan^{-1}(x)$, looks and reads way nicer!!