# How do you differentiate the following parametric equation:  x(t)=lnt/t, y(t)=(t-3)^2 ?

Mar 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {t}^{2} \left(t - 3\right)}{1 - \ln t} , t \ne 0 , t \ne e .$

#### Explanation:

By the Rule of Parametric Diffn., $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} , x ' \left(t\right) \ne 0.$

Now, using the Chain Rule,

$y ' \left(t\right) = \frac{d}{\mathrm{dt}} \left\{{\left(t - 3\right)}^{2}\right\} = 2 \left(t - 3\right) \frac{d}{\mathrm{dt}} \left\{\left(t - 3\right)\right\} .$

$\therefore y ' \left(t\right) = 2 \left(t - 3\right) \ldots \ldots . . \left(1\right) .$

Next, $x \left(t\right) = \ln \frac{t}{t}$

$\therefore x ' \left(t\right) = \frac{t \frac{d}{\mathrm{dt}} \left(\ln t\right) - \left(\ln t\right) \frac{d}{\mathrm{dt}} \left(t\right)}{t} ^ 2. \ldots \text{[The Quotient Rule]}$

$= \frac{t \left(\frac{1}{t}\right) - \left(\ln t\right) \left(1\right)}{t} ^ 2$

$\therefore x ' \left(t\right) = \frac{1 - \ln t}{t} ^ 2. \ldots \ldots \ldots \left(2\right) .$

Also, let us note that,

$x ' \left(t\right) = 0 \Rightarrow t = e , \mathmr{and} , t \ne e \Rightarrow x ' \left(t\right) \ne 0. \ldots \ldots . . \left(2 '\right) .$

Altogether, taking into account $\left(1\right) , \left(2\right) , \mathmr{and} \left(2 '\right) ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(t - 3\right)}{\frac{1 - \ln t}{t} ^ 2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {t}^{2} \left(t - 3\right)}{1 - \ln t} , t \ne 0 , t \ne e .$

Enjoy Maths.!