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

Dec 22, 2016

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

#### Explanation:

differentiate both side wrt $x$

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

points to note:

1) When differentiating $f \left(y\right)$ wrt $\text{x}$ differentiate wrt $y$ and then multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$

2) for the first term on RHS the product rule needs to be employed.

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

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

Now rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \left(\ln y + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 1$

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

Dec 22, 2016

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

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

To differentiate $y \ln y \text{ use " color(blue)"product rule}$

$0 = \left(y \times \frac{1}{y} . \frac{\mathrm{dy}}{\mathrm{dx}} + \ln y \times 1. \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 1$

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

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