Question

How do you differentiate the following parametric equation: `x(t)=e^t/(t+t), y(t)=t-e^(3t)`?

Answer 1

`dy/dx = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1)`

Explanation:

I assume that you want more than just the derivatives of the parametric equation, and actually require `dy/dx`

We have;

`x(t) = e^t/(t+t) = e^t/(2t)`

Differentiating wrt `t`, and applying the quotient rule, we have;

`dx/dt = {(2t)(e^t) - (2)(e^t)}/(2t)^2`
`" " = (2(t-1)e^t)/(4t^2)`
`" " = ((t-1)e^t)/(2t^2)`

And:

`y(t)=t-e^(3t)`

Differentiating wrt `t`, and applying the chain rule, we have;

`dy/dt = 1 - 3e^(3t)`

And finally by the chain rule:

`dy/dx = (dy//dt) / (dx//dt)`
`" " = (1 - 3e^(3t)) / (((t-1)e^t)/(2t^2))`
`" " = (1 - 3e^(3t)) * (2t^2)/((t-1)e^t)`
`" " = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1)`