Let $\displaystyle x=t^2+t$, and $\displaystyle y=sin(t)$
Find $\displaystyle d/dx(dy/dx)$
Thanks
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)
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}} = $
$\displaystyle \frac{x'(t) \cdot y''(t) - y'(t) \cdot x''(t)}{[x'(t)]^3} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}$