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

Jun 28, 2016

The derivative of $f \left(t\right) = f ' \left(t\right) = \left(\frac{2}{t} \cdot \ln t - 1 , \sec t \left(t \cdot \tan t + 1\right)\right) .$

#### Explanation:

Let $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right) ,$ so that $f ' \left(t\right) = \left(x ' \left(t\right) , y ' \left(t\right)\right) .$

Now, $x \left(t\right) = {\left(\ln t\right)}^{2} - t \Rightarrow x ' \left(t\right) = \frac{d}{\mathrm{dt}} \left\{{\left(\ln t\right)}^{2} - t\right\} = \frac{d}{\mathrm{dt}} {\left(\ln t\right)}^{2} - \frac{d}{\mathrm{dt}} \left(t\right) = 2 \ln t \cdot \frac{d}{\mathrm{dt}} \left(\ln t\right) - 1 = \frac{2}{t} \cdot \ln t - 1.$

$y \left(t\right) = t \sec t \Rightarrow y ' \left(t\right) = t \cdot \frac{d}{\mathrm{dt}} \left(\sec t\right) + \sec t \cdot \frac{d}{\mathrm{dt}} \left(t\right) = t \cdot \sec t \cdot \tan t + \sec t = \sec t \left(t \cdot \tan t + 1\right) .$

Hence, the derivative of $f \left(t\right) = f ' \left(t\right) = \left(\frac{2}{t} \cdot \ln t - 1 , \sec t \left(t \cdot \tan t + 1\right)\right) .$