# What is the derivative of f(t) = (tlnt, 3t^2+5t ) ?

Oct 8, 2017

$f ' \left(t\right) = \frac{6 t + 5}{1 + \ln t}$

#### Explanation:

We have a parametric function of two variables, $x$ and $y$, say:

$f \left(t\right) = \left\{\begin{matrix}\setminus x \left(t\right) & = t \ln t \\ \setminus y \left(t\right) & = 3 {t}^{2} + 5 t\end{matrix}\right.$

Differentiating wrt $t$ we have (via the product rule):

$\frac{\mathrm{dx}}{\mathrm{dt}} = t \left(\frac{d}{\mathrm{dt}} \ln t\right) + \left(\frac{d}{\mathrm{dt}} t\right) \ln t$
$\setminus \setminus \setminus \setminus \setminus = t \left(\frac{1}{t}\right) + \left(1\right) \ln t$
$\setminus \setminus \setminus \setminus \setminus = 1 + \ln t$

And:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 6 t + 5$

Finally, by the chain rule:

$f ' \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{6 t + 5}{1 + \ln t}$