How do you use implicit differentiation to find dy/dx given ln(x-2)=ln(2y+1)?

Feb 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

Explanation:

We don't actually need to use implicit differentiation (at least in the logarithmic form) due to the fact that:

$\ln A = \ln B \iff A = B$

So if:

$\ln \left(x - 2\right) = \ln \left(2 y + 1\right) \iff x - 2 = 2 y + 1$

$\therefore 2 y = x - 3$
$\therefore \setminus \setminus y = \frac{1}{2} x - \frac{3}{3}$

And so if we differentiate we get;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

If you are not convinced by that or just just love Calculus then we can differentiate the initial equation implicitly:

$\ln \left(x - 2\right) = \ln \left(2 y + 1\right) \iff x - 2 = 2 y + 1$

$\therefore \frac{1}{x - 2} = \frac{1}{2 y + 1} \cdot 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \frac{2}{2 y + 1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x - 2}$

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

Now at a first glance this seems very different to the answer above, but is in fact the same, which we can show as follows:

Take Natural Logarithms of both sides:

$\ln \left\{2 \frac{\mathrm{dy}}{\mathrm{dx}}\right\} = \ln \left\{\frac{2 y + 1}{x - 2}\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln \left(2 y + 1\right) - \ln \left(x - 2\right) \setminus \setminus \setminus$ (rule of logs)

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 \setminus \setminus \setminus$ (from the initial equation)

$\therefore 2 \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{0}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \setminus \setminus \setminus$ (as above)