# What is the derivative of f(t) = (t/(t+1) , 1/(t^2-t) ) ?

Jan 20, 2018

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

#### Explanation:

We know that $x = \frac{t}{t + 1}$ and $y = \frac{1}{{t}^{2} - t}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \div \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\left({t}^{2} - t\right) \frac{d}{\mathrm{dt}} \left[1\right] - 1 \frac{d}{\mathrm{dt}} \left[{t}^{2} - t\right]}{{t}^{2} - t} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dt}}} = \frac{0 \left({t}^{2} - t\right) - 1 \left(2 t - 1\right)}{{t}^{2} - t} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dt}}} = \frac{- 2 t + 1}{{t}^{2} - t} ^ 2$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\left(t + 1\right) \frac{d}{\mathrm{dt}} \left[t\right] - t \frac{d}{\mathrm{dt}} \left[t + 1\right]}{t + 1} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{1 \left(t + 1\right) - t \left(1\right)}{t + 1} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{t + 1 - t}{t + 1} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{1}{t + 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{- 2 t + 1}{{t}^{2} - t} ^ 2}{\frac{1}{t + 1} ^ 2}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{{\left(t + 1\right)}^{2} \left(- 2 t + 1\right)}{{t}^{2} - t} ^ 2$