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

Aug 3, 2016

Hence derivative of $f \left(t\right) = \left(\frac{t}{t + 1} , t {e}^{2 t - 1}\right)$ is ${\left(t + 1\right)}^{2} \left(1 + 2 t\right) {e}^{2 t - 1}$

#### Explanation:

In parametric form of function where $f \left(t\right) = \left(g \left(t\right) , h \left(t\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dh}}{\mathrm{dt}}}{\frac{\mathrm{dg}}{\mathrm{dt}}}$

as such derivative of given function $f \left(t\right) = \left(\frac{t}{t + 1} , t {e}^{2 t - 1}\right)$ is

(d/dt(te^(2t-1)))/(d/(dt)(t/(t+1))

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

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

Hence derivative of $f \left(t\right) = \left(\frac{t}{t + 1} , t {e}^{2 t - 1}\right)$ is

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