Derivative of a parametric equation

[Karol]

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]

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 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 $y$.

 

[Karol]

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]

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.

 

[Karol]

Thank you fresh_42, Dick and Mark44

 

[Ray Vickson]

Why? indeed i cannot express t as a function of x, is that the reason?

You have already been told (in another, similar, thread) that solving for $t$ in terms of $x$ will give you two different formulas, so will give you two different curves $y = f_1(x)$ and $y = f_2(x)$. That means you will get two different derivative formulas for $dy/dx$.