# Parametric Form of the Derivative

## First Derivative

Let's say you have the parametric equation x = f(t), and y = g(t). dy/dx, or the combined equation (if you were to merge the two equations into one) equals (dy/dt)/(dx/dt) or g'(t)/f'(t).

### Proof

If you let Δy = g(t + Δt) - g(t), and Δx = f(t + Δt), you can say

dy/dx = lim as Δx->0 (Δy/Δx) = lim[ (g(t + Δt) - g(t))/(f(t + Δt) - f(t))]

If you divide both the numerator and denominator by Δx, you get lim [(g(t + Δt) - g(t))/Δt] / lim [(f(t + Δt) - f(t)/Δt].

Recall that that's the definition of a derivative, so you get dy/dx = g'(t)/f'(t) = (dy/dt)/(dx/dt).

(Proof from textbook)

## Second and Higher Degree Derivatives

The second derivative is the *derivative of the derivative / derivative of x *, so it's:

The 3rd,4th, 5th, etc. derivatives are almost the same as the second derivative. In the third derivative, you replace dy/dx with the second derivative. The idea is the same, you want to find the derivative of that, but you must divide by the derivative of x, because x changes differently than usual (if x = t^2, it's not a linear change).

# Example

x = √t, y = 1/4(t^{2} - 4), t≥0

Find the derivative and second derivative at (2,3)

dy/dx = (dy/dt)/(dx/dt) =

## (t/2) /(1/(2√t) = t√t = t^{3/2}

Derivative at (2,3): x = 2, t = 4, 4^{3/2} = **8**

Second derivative = d/dt [t^{3/2}]/(dx/dt) = (3/2)√t/(1/(2√t)) = 3t

When t = 4, 3t = 12.

So, the graph is concave up at (2,3), and the slope is 8.

# Speed

Sometimes, a problem might ask what the speed of a set of parametric equations is.

## Example

# Velocity

Sometimes, a problem might ask about the velocity.