Find dy/dx for x=1/t+1 and y=1/t-1

Answer 1

To find dy/dx, we can first find dy/dt and dx/dt, and then use the chain rule.

To find dy/dt, we can differentiate the equation y = 1/t - 1 with respect to t:
dy/dt = d/dt(1/t - 1)
= -1/t^2

To find dx/dt, we can differentiate the equation x = 1/(t+1) with respect to t:
dx/dt = d/dt(1/(t+1))
= -1/(t+1)^2

Now, we can apply the chain rule to find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
= (-1/t^2) / (-1/(t+1)^2)
= (t+1)^2 / t^2
= (t^2 + 2t + 1) / t^2

So, dy/dx = (t^2 + 2t + 1) / t^2.

Answer 2

Is it
Find dy/dx for x=1/t+1 and y=1/t-1
or is it
Find dy/dx for x=1/ (t+1) and y=1/ (t-1)

Answer 3

Apologies for the confusion. The equations are:

x = 1/(t + 1) and y = 1/(t - 1)

To find dy/dx, we will take the derivatives of y and x with respect to t, and then apply the chain rule.

First, let's differentiate y with respect to t:
dy/dt = d/dt(1/(t - 1))
dy/dt = -1/(t - 1)^2

Now, let's differentiate x with respect to t:
dx/dt = d/dt(1/(t + 1))
dx/dt = -1/(t + 1)^2

Next, we can use the chain rule to find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
dy/dx = (-1/(t - 1)^2)/(-1/(t + 1)^2)
dy/dx = (t + 1)^2/(t - 1)^2

Therefore, dy/dx = (t + 1)^2/(t - 1)^2.