# Derivative of a Parametic Function

How to find the derivative of a parametric function: 1 example and its solution.

## Examplex = t^{3} - 2t, y = t^{2} + 1, dy/dx = ?

A parametric function is a function

that a parameter (t) connects the variables (x, y).

In this example,

x and y are not directly connected by a formula: like y = f(x).

But the parameter t connects both x and y.

Let's see how to find dy/dx

at t = 1.

Find dx/dt.

x = t^{3} - 2t

So dx/dt = 3t^{2} - 2.

Derivative of a Polynomial

Find dy/dt.

y = t^{2} + 1

So dy/dt = 2t^{1} + 0 = 2t.

It says to find dy/dx.

But you found dx/dt and dy/dt.

So change dy/dx to (dy/dt)/(dx/dt).

It's like dividing both of the numerator and the denominator by dt.

dy/dt = 2t

dx/dt = 3t^{2} - 2

Then (dy/dt)/(dx/dt) = 2t/(3t^{2} - 2).

Don't put dx/dt = 3t^{2} - 2 in the numerator.

It's easy to make this mistake.

It says to find dy/dx at t = 1.

So put t = 1 into dy/dx = 2t/(3t^{2} - 2).

Then [dy/dx]_{t = 1} = 2⋅1/(3⋅1^{2} - 2).

Then you get 2.

So dy/dx = 2

at t = 1.