# Derivative of a parametric equation

## Homework Statement

$$y=1+t^2,~~x=\frac{t}{1+t^2}$$
What is dy/dx

## Homework Equations

Parametric equation's derivative:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

## The Attempt at a Solution

$$\frac{dx}{dt}=\frac{1-t^2}{(1+t^2)^2}$$
$$\frac{dy}{dx}=\frac{2t(1+t^2)^2}{1-t^2}$$
I can't translate it back to x

fresh_42
Mentor
Why do you want to do it? Anyway, you could also start from ##t=x\cdot y## and express ##\frac{dy}{dx}## as a function of ##x## and ##y##.

Dick
Homework Helper

## Homework Statement

$$y=1+t^2,~~x=\frac{t}{1+t^2}$$
What is dy/dx

## Homework Equations

Parametric equation's derivative:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

## The Attempt at a Solution

$$\frac{dx}{dt}=\frac{1-t^2}{(1+t^2)^2}$$
$$\frac{dy}{dx}=\frac{2t(1+t^2)^2}{1-t^2}$$
I can't translate it back to x

The derivative won't be a function only of ##x##. Use that ##t=xy## if you want to express it as a function of ##x## and ##y##.

The derivative won't be a function only of ##x##.
Why? indeed i cannot express t as a function of x, is that the reason?

Mark44
Mentor
Why? indeed i cannot express t as a function of x, is that the reason?
Right. To solve for t in the equation ##x = \frac t {1 + t^2}##, you would most likely use the quadratic formula, which will give two values of t. So t is not a function of x.

Thank you fresh_42, Dick and Mark44

Ray Vickson