# What is the derivative of f(t) = (2t-3te^t, 2t^2+3t ) ?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 t + 3}{2 - 3 t {e}^{t} - 3 {e}^{t}}$

#### Explanation:

Given $f \left(t\right) = \left(2 t - 3 t {e}^{t} , 2 {t}^{2} + 3 t\right)$

$x = 2 t - 3 t {e}^{t}$ and $y = 2 {t}^{2} + 3 t$

solve for $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Solve for $\frac{\mathrm{dx}}{\mathrm{dt}}$

$x = 2 t - 3 t {e}^{t}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 - 3 \left[t \cdot {e}^{t} + {e}^{t} \cdot 1\right] = 2 - 3 t {e}^{t} - 3 {e}^{t}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$

$y = 2 {t}^{2} + 3 t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t + 3$

Solve now for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{4 t + 3}{2 - 3 t {e}^{t} - 3 {e}^{t}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 t + 3}{2 - 3 t {e}^{t} - 3 {e}^{t}}$

God bless...I hope the explanation is useful.