1. ## Tricky Derivative

Hi, I have the following question:
Given that $x = e^{t}$, find dy/dx in terms of dy/dt and show that

$\frac{d^{2}y}{dx^{2}} = e^{-2t}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt})$

So first of all, I used the chain rule to get

dy/dx = dt/dx . dy/dt

dt/dx = $e^{-t}$

so that gives me $\frac{dy}{dx} = e^{-t}\frac{dy}{dt}$
but how do I differentiate that a second time. I would have to be the product rule, but it's not implicit? I can't see how

Thanks

2. Originally Posted by thomas49th
Hi, I have the following question:
Given that $x = e^{t}$, find dy/dx in terms of dy/dt and show that

$\frac{d^{2}y}{dx^{2}} = e^{-2t}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt})$

So first of all, I used the chain rule to get

dy/dx = dt/dx . dy/dt

dt/dx = $x = e^{-t}$

so that gives me $\frac{dy}{dx} = e^{-t}\frac{dy}{dt}$
but how do I differentiate that a second time. I would have to be the product rule, but it's not implicit? I can't see how

Thanks
$\frac{d}{dx} \left[ e^{-t}\frac{dy}{dt} \right] = \frac{d}{dt} \left[ e^{-t} \, \frac{dy}{dt} \right] \cdot \frac{dt}{dx}$

$= \left( -e^{-t} \, \frac{dy}{dt} + e^{-t} \, \frac{d^2 y}{dt^2}\right) \cdot e^{-t}$.