# Proving 2nd ode [closed]

If $$x=e^t$$, can someone give me proof that $$\frac{d^2}{dx^2}=\frac{1}{e^{2t}}\left(\frac{d^2}{dt^2}−\frac{d}{dt}\right).$$ Thank you

## closed as off-topic by caverac, RRL, Saad, José Carlos Santos, mrtaurhoDec 27 '18 at 12:35

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$$\frac{d}{dx} = \frac{d}{dt} \times \frac{dt}{dx} = \frac{d/dt}{dx/dt} = e^{-t} \frac{d}{dt}.$$

Note that $$\frac{d^2}{dx^2} = \frac{d}{dx} \left[ \frac{d}{dx} \right]$$ and you can now use the previous result.