1. ## Parametric Derivative Question

Let x=t2 + t, and let y = sin t.

a. Find d/dt (dy/dx) as a function of t.

b. Find d/dx (dy/dx) as a function of t.

I understand how to do (a) part. Using the rule dy/dx = (dy/dt)/(dx/dt), I get dy/dx = (cos t) /(2t + 1). Using the Quotient Rule,
I get d/dt (dy/dx) = [-(2t +1)sin t - 2 cos t] / (2t +1)2.

How do I find d/dx (dy/dx)? I think it would involve the Chain Rule but not sure where to start.

2. ## Re: Parametric Derivative Question

We know that

$\dfrac d {dx} y = \dfrac {dy}{dx}=\dfrac {\dfrac {dy}{dt}}{\dfrac {dx}{dt}}$

so it stands to reason that

$\dfrac d {dx} \dfrac {dy}{dx} = \dfrac {\dfrac d {dt} \dfrac {dy}{dx}}{\dfrac {dx}{dt}}=$

$-\dfrac {\sin(2t+1)+2\cos(t)}{(2t+1)^3}$