# How do you implicitly differentiate 6=ylny/x?

Sep 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{1 + \ln y} , \mathmr{and} , = 6 {\left(\ln e y\right)}^{-} 1$.

#### Explanation:

Rewriting the eqn. as, $: 6 x = y \ln y$

$\therefore \ln \left(6 x\right) = \ln \left(y \ln y\right)$

$\therefore \ln 6 + \ln x = \ln y + \ln \left(\ln y\right)$

$\therefore \frac{d}{\mathrm{dx}} \ln 6 + \ln x = \frac{d}{\mathrm{dx}} \left(\ln y + \ln \left(\ln y\right)\right)$

$\therefore 0 + \frac{1}{x} = \frac{d}{\mathrm{dy}} \left(\ln y + \ln \left(\ln y\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} \ldots \ldots \ldots \text{[Chain Rule]}$

$\therefore \frac{1}{x} = \left\{\frac{1}{y} + \frac{1}{\ln} y \frac{d}{\mathrm{dy}} \left(\ln y\right)\right\} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \frac{1}{x} = \left\{\left(\frac{1}{y} + \frac{1}{\ln} y \cdot \frac{1}{y}\right)\right\} \frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \ln y}{x \left(1 + \ln y\right)}$.

Remembering that, by the original eqn., $y \ln y = 6 x$, we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6}{1 + \ln y} , \mathmr{and} , \frac{6}{\ln e + \ln y} = \frac{6}{\ln} \left(e y\right) = 6 {\left(\ln e y\right)}^{-} 1$.

Alternatively,

$6 x = y \ln y \Rightarrow \frac{d}{\mathrm{dy}} \left(6 x\right) = \frac{d}{\mathrm{dy}} \left(y \ln y\right)$

$\Rightarrow 6 \frac{\mathrm{dx}}{\mathrm{dy}} = y \frac{d}{\mathrm{dy}} \left(\ln y\right) + \left(\ln y\right) \cdot \frac{d}{\mathrm{dy}} \left(y\right)$

$\Rightarrow 6 \frac{\mathrm{dx}}{\mathrm{dy}} = y \cdot \frac{1}{y} + \ln y = 1 + \ln y$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1 + \ln y}{6}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \frac{6}{1 + \ln y}$, as before.!

Enjoy maths.!