# How do we implicitly differentiate xlny+ylnx=2?

Mar 1, 2017

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

#### 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 y are written implicitly as functions of $x$. So 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 $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Hence differentiating $x \ln y + y \ln x = 2$

$1 \times \ln y + x \times \frac{1}{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} \ln x + y \times \frac{1}{x} = 0$

or $\ln y + \frac{x}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \ln x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{y}{x} = 0$

or $\left(\frac{x}{y} + \ln x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(\frac{y}{x} + \ln y\right)$

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