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

Nov 7, 2016

$f ' \left(t\right) = \frac{6 t - 1}{1 + \ln t}$

Explanation:

Let $\left\{\begin{matrix}x \left(t\right) = t \ln t \\ y \left(t\right) = 3 {t}^{2} - t\end{matrix}\right.$

Then differentiating wrt $t$ (using the product rule)

$\left\{\begin{matrix}\frac{\mathrm{dx}}{\mathrm{dt}} = \left(t\right) \left(\frac{1}{t}\right) + \left(1\right) \ln t & = 1 + \ln t \\ \frac{\mathrm{dy}}{\mathrm{dt}} & = 6 t - 1\end{matrix}\right.$

By the chain rule, we have
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$, or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 t - 1}{1 + \ln t}$