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

Apr 5, 2018

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1 - {e}^{t}}{2 t - \cos t}$

#### Explanation:

Derivative of $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $x \left(t\right) = {t}^{2} - \sin t$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - \cos t$

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

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1 - {e}^{t}}{2 t - \cos t}$