Find d y d x if : x^2+e^ x y =y^2+ (x+y)

Question

Find $\frac{d y}{d x}$ if : $x^2+e^{x y}=y^2+\log (x+y)$

✓

Answer

Given that : $x^2+e^{x y}=y^2+\log (x+y)$
Recall that $\frac{d}{d x} g(f(x))=g^{\prime}(f(x)) \cdot \frac{d}{d x} f(x)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left[e^{x y}\right]=\frac{d}{d x}\left(y^2\right)+\frac{d}{d x}[\log (x+y)] \\
& 2 x+e^{x y} \frac{d}{d x}(x y)=2 y \frac{d y}{d x}+\frac{1}{x+y} \frac{d}{d x}(x+y) \\
& 2 x+e^{x y}\left[x \frac{d y}{d x}+y(1)\right]=2 y \frac{d y}{d x}+\frac{1}{x+y}\left[1+\frac{d y}{d x}\right] \\
& 2 x+x e^{x y} \frac{d y}{d x}+y e^{x y}=2 y \frac{d y}{d x}+\frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=2 y \frac{d y}{d x}-x e^{x y} \frac{d y}{d x}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=\left[2 y-x e^{x y}+\frac{1}{x+y}\right] \frac{d y}{d x} \\
& \frac{2 x(x+y)+y e^{x y}(x+y)-1}{x+y}=\left[\frac{2 y(x+y)-x e^{x y}(x+y)+1}{x+y}\right] \frac{d y}{d x} \\
\therefore & \frac{d y}{d x}=\frac{2 x(x+y)+y e^{y y}(x+y)-1}{2 y(x+y)-x e^{y y}(x+y)+1}
\end{aligned}
$

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