# If x+y-1=ln(x^2+y^2), what is (dy)/(dx)?

Oct 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + {y}^{2} - 2 x}{2 y - {x}^{2} - {y}^{2}}$

#### Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$.

However, some functions like the one given above, are written implicitly as functions of $x$ and $y$. Here, what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Thus as $x + y - 1 = \ln \left({x}^{2} + {y}^{2}\right)$, differentiating w.r.t. $x$, we have

$1 + \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} + {y}^{2}} \left(2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

or $1 - \frac{2 x}{{x}^{2} + {y}^{2}} = \frac{2 y}{{x}^{2} + {y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}}$

or $1 - \frac{2 x}{{x}^{2} + {y}^{2}} = \left[\frac{2 y}{{x}^{2} + {y}^{2}} - 1\right] \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\frac{{x}^{2} + {y}^{2} - 2 x}{{x}^{2} + {y}^{2}} = \left[\frac{2 y - {x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}\right] \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + {y}^{2} - 2 x}{2 y - {x}^{2} - {y}^{2}}$