I want to find an expression for the derivative of the implicit function below:
$$\arctan(xy)=\frac{\pi}{4}e^{x-y}, \qquad \text{at point } (1,1)$$
I've tried to derive this using both Maple and Wolfram Alpha, and it appears that both of them give different results. So, I tried two different techniques (by hand):
1) Technique 1 (same result with Wolfram Alpha): differentiating both sides:$$\frac{d}{dx}\arctan(xy)=\frac{d}{dx}\frac{\pi}{4}e^{x-y}\\ \frac{y+xy'}{1+x^2y^2} = \frac{\pi}{4}(1-y')e^{x-y}\\ \frac{1+y'}{2} = \frac{\pi}{4}(1-y')e^{0}\\ 2(1+y')=\pi(1-y')\\ 2+2y'=\pi-\pi y'\\ y'=\frac{\pi-2}{\pi+2} $$
2) Technique 2 (same result with Maple): expressing as $f(x,y)=C, \quad C \text{ is a constant}$
$$\frac{\arctan(xy)}{e^{x-y}}=\frac{\pi}{4}\\ \frac{\frac{y+xy'}{1+x^2y^2}e^{x-y}- \arctan(xy)(1-y')e^{x-y}}{(e^{x-y})^2} = 0\\ \frac{\frac{1+y'}{2}e^{0}- \arctan(1)(1-y')e^{0}}{(e^{o})^2} = 0\\ (1+y')- 2\arctan(1)(1-y') = 0\\ y'=\frac{2\arctan(1)+1}{2\arctan(1)+1} = 1 $$
Which one of these is right?
Edit: I think I see the error in technique 2. However, Maple insists it's $1$. $$1+y'-\frac{\pi}{2}(1-y')=0\\ 2+2y'-\pi+\pi y'=0\\ y'(2+\pi) +2-\pi = 0\\ y'= \frac{\pi-2}{\pi+2} $$