Find $y'$ for $\ln(x+y)=\arctan(xy)$

Question

Find $y'$ for $\ln(x+y)=\arctan(xy)$

Here is my attempt at a solution. Is this correct? Any hints or advice would be appreciated.

Answer 1

Even if the result is correct, I guess that a systematic use of total differentiation makes things slightly easier.

Consider $$F=\log(x+y)-\tan^{-1}(xy)=0$$ So $F'_x~dx+F'_y~dy=0$. I this case, $$F'_x=\frac{1}{x+y}-\frac{y}{x^2 y^2+1}$$ $$F'_y=\frac{1}{x+y}-\frac{x}{x^2 y^2+1}$$ $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{\frac{y}{x^2 y^2+1}-\frac{1}{x+y}}{\frac{1}{x+y}-\frac{x}{x^2 y^2+1}}=\frac{y \left(x+y-x^2y\right)-1}{x^2 \left(y^2-1\right)-x y+1}=- \frac{(x+y)y-(1+x^2y^2)}{(x+y)x-(1+x^2y^2)}$$

Answer 2

whilst your answer cannot be improved on, it may be of interest to look at a parametric approach. if we set: $\xi $ to be the common value of $\log(x+y)$ and $\arctan(xy)$ then $x$ and $y$ are the two roots of: $$ z^2 -e^{\xi}z + \tan \xi =0 $$ differentiating with respect to $\xi$ gives: $$ 2zz' - e^{\xi}(z+z') +\sec^2 \xi =0 $$ or $$ z' (2z-x-y) = (x+y)z - (1+x^2y^2) $$ if we substitute the roots $x$ and $y$ for $z$ we obtain: $$ z_y' (y-x) = (x+y)y-(1+x^2y^2) \\ z_x' (x-y) = (x+y)x-(1+x^2y^2) $$ hence: $$ \frac{dy}{dx} = \frac{z_y'}{z_x'} = - \frac{(x+y)y-(1+x^2y^2)}{(x+y)x-(1+x^2y^2)} $$