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

May 7, 2017

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

#### Explanation:

This is a parametric form of equation i.e. $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

In such a case $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

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

and $y \left(t\right) = {e}^{2 t - 1}$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 {e}^{2 t - 1}$

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

May 7, 2017

For a parametric function $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, the function's derivative is just given by the $x$ and $y$ components' derivatives: $f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right)$.

So:

$f \left(t\right) = \left(- {\left(t + 1\right)}^{-} 1 , {e}^{2 t - 1}\right)$

$f ' \left(t\right) = \left(- \left(- {\left(t + 1\right)}^{-} 2\right) , 2 {e}^{2 t - 1}\right)$

$f ' \left(t\right) = \left(\frac{1}{t + 1} ^ 2 , 2 {e}^{2 t - 1}\right)$