# Derivative of a Parametric Function

• Oct 5th 2008, 07:26 AM
nic42991
Derivative of a Parametric Function
Let $\displaystyle x=t^2+t$, and $\displaystyle y=sin(t)$

Find $\displaystyle d/dx(dy/dx)$

Thanks
• Oct 5th 2008, 07:46 AM
skeeter
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\displaystyle \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}$
• Oct 5th 2008, 08:25 AM
nic42991
I don't understand how that works out. I understand $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, but I don't understand this part:

http://www.mathhelpforum.com/math-he...4bc050eb-1.gif

I don't have the intuition behind it.
• Oct 5th 2008, 08:39 AM
Moo
Quote:

Originally Posted by nic42991
I don't understand how that works out. I understand $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, but I don't understand this part:

http://www.mathhelpforum.com/math-he...4bc050eb-1.gif

I don't have the intuition behind it.

If you prefer, you can say that :

$\displaystyle \frac{df}{dx}=\frac{\frac{df}{dt}}{\frac{dx}{dt}}$ (1)

Now, let $\displaystyle f=\frac{dy}{dx}$ in (1), this gives :

$\displaystyle \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}$

And then transform $\displaystyle \frac{dy}{dx}$, by letting f=y in (1)
• Oct 5th 2008, 08:43 AM
skeeter
let's look at it another way ...

let $\displaystyle \frac{dy}{dt} = y'(t)$ and $\displaystyle \frac{dx}{dt} = x'(t)$

$\displaystyle \frac{d}{dx}\left[\frac{y'(t)}{x'(t)}\right] =$

using the quotient and chain rule ...

$\displaystyle \frac{x'(t) \cdot y''(t) \cdot \frac{dt}{dx} - y'(t) \cdot x''(t) \cdot \frac{dt}{dx}}{[x'(t)]^2} =$

$\displaystyle \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^2 \cdot \frac{dx}{dt}} =$

$\displaystyle \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^3} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}$