Question

What is the derivative of `f(t) = (-1/(t+1) , e^(2t-1) )`?

Answer 1

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

Explanation:

This is a parametric form of equation i.e. `f(t)=(x(t),y(t))`

In such a case `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`

Here `x(t)=-1/(t+1)` and `(dx)/(dt)=-(-1)/(t+1)^2=1/(t+1)^2`

and `y(t)=e^(2t-1)` and `(dy)/(dt)=2e^(2t-1)`

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

Answer 2

For a parametric function `f(t)=(x(t),y(t))`, the function's derivative is just given by the `x` and `y` components' derivatives: `f'(t)=(x'(t),y'(t))`.

So:

`f(t)=(-(t+1)^-1,e^(2t-1))`

`f'(t)=(-(-(t+1)^-2),2e^(2t-1))`

`f'(t)=(1/(t+1)^2,2e^(2t-1))`