# How do you find the derivative of g(t) = (ln(kt)+2t) / (ln(kt)-2t)?

Jun 3, 2015

Here, we'll use the quotient rule, which states that, be $y = f \frac{x}{g} \left(x\right)$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

Let's just find the derivatives and use the function properly:

• $f \left(x\right) = \ln k t + 2 t$

• $f ' \left(x\right)$ will demand chain rule, which states: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Thus, renaming $u = k t$, we have $f ' \left(x\right) = \left(\frac{1}{u}\right) \left(k\right) + 2 = \left(\frac{\cancel{k}}{\cancel{k} t}\right) + 2 = \frac{1}{t} + 2$

• $g \left(x\right) = \ln k t - 2 t$

• $g ' \left(x\right)$ follows the logic for $f ' \left(x\right)$: $g ' \left(x\right) = \frac{1}{t} - 2$

Now, applying the quotient rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\frac{1}{t} + 2\right) \left(\ln k t - 2 t\right) - \left(\ln k t + 2 t\right) \left(\frac{1}{t} - 2\right)}{\ln k t - 2 t} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cancel{\frac{\ln k t}{t}} - 2 + 2 \ln k t \cancel{- 4} t \cancel{- \frac{\ln k t}{t}} + 2 \ln k t - 2 \cancel{+ 4 t}}{\ln k t - 2 t} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \ln k t - 4}{\ln k t - 2 t} ^ 2 = \frac{4 \left(\ln k t - 1\right)}{\ln k t - 2 t} ^ 2$