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

Mar 1, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{t \sin t + {t}^{2} \cos t}{2 \ln t - t}$

#### Explanation:

Derivative of a parametric function such as $f \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$.

Here as $y \left(t\right) = t \sin t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 \times \sin t + t \times \cos t = \sin t + t \cos t$

and as $x \left(t\right) = {\left(\ln t\right)}^{2} - t$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \ln t \times \frac{1}{t} - 1 = \frac{2 \ln t - t}{t}$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\sin t + t \cos t}{\frac{2 \ln t - t}{t}} = \frac{t \sin t + {t}^{2} \cos t}{2 \ln t - t}$