0

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.

Ramified_Minds
  • 709
Rsherrill
  • 1
  • do you mean $f(x,y)=y\tan^{-1}(xy)$? (i prefer the notion $\arctan(x)$ for $\tan(x)$ – slinshady Apr 15 '15 at 21:44
  • 1
    Pretend you have $f(x)=6 \tan^{-1}(6x)$ , could you find $f'(x)$? – randomgirl Apr 15 '15 at 21:48
  • As a warmup problem, can you find the derivative of $ g (x) = 5 \tan^{-1}(5x) $? – littleO Apr 15 '15 at 21:48
  • Differentiate once with respect to $x$, then once with respect to $y$. When doing this, treat the other variable as if it were a constant. For example, when first differentiating with respect to $x$, it's much like trying to compute $$\frac{d}{dx}(7\tan^{-1} 7x)$$ Don't really put numbers in, of course, but think of the other variable that way. – MPW Apr 15 '15 at 21:48

2 Answers2

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.

Ramified_Minds
  • 709
0

$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!!

CivilSigma
  • 1,047
  • Can you walk me through the process of determining the first partial derivative of x? I'm not understanding where the y^2 is coming from. – Rsherrill Apr 15 '15 at 21:50
  • When taking the partial derivative with respect to x, you treat all other variables as constants. What happens when you differentiate $arctan(xy)$? you get $\frac{y}{1+x^2y^2}$. Then $f_x=y * \frac{y}{1+x^2y^2}$ – CivilSigma Apr 15 '15 at 21:52