# What is the derivative of f(t) = (t/(t+1) , t/(4t^2-t) ) ?

Sep 28, 2016

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

#### Explanation:

$f \left(t\right) = \left(\frac{t}{t + 1} , \frac{t}{4 {t}^{2} - t}\right)$ is parametric form of equation,

where $\frac{t}{t + 1}$ represents $x$ and $\frac{t}{4 {t}^{2} - t}$ represents $y$.

In such cases, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here we can find $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ using quotient rule

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1 \times \left(4 {t}^{2} - t\right) - t \times \left(8 t - 1\right)}{4 {t}^{2} - t} ^ 2 = \frac{4 {t}^{2} - t - 8 {t}^{2} + t}{4 {t}^{2} - t} ^ 2 = \frac{- 4 {t}^{2}}{4 {t}^{2} - t} ^ 2$

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

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 {t}^{2}}{4 {t}^{2} - t} ^ 2 \div \frac{1}{t + 1} ^ 2$

= $\frac{- 4 {t}^{2} {\left(t + 1\right)}^{2}}{4 {t}^{2} - t} ^ 2 = \frac{- 4 {t}^{2} {\left(t + 1\right)}^{2}}{{t}^{2} {\left(4 t - 1\right)}^{2}} = \frac{- 4 {\left(t + 1\right)}^{2}}{4 t - 1} ^ 2$