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If $x=e^t$, show that (using the chain rule) $$\frac{d}{dx} = e^{-t}\frac{d}{dt}$$ and $$\frac{d^2}{dx^2} = e^{-2t}\left[\frac{d^2}{dt^2}-\frac{d}{dt}\right].$$

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    If $x=e^x$, $x$ is a constant. Perhaps you meant $x=e^{t}$. – 2017-02-10
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    @tilper $\frac{d}{dx}$ is an operator. – 2017-02-10
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    $\frac{d}{dx}=\frac{d}{dt} \frac{dt}{dx}$ by chain rule – 2017-02-10
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    I did mean e^(t). – 2017-02-10

2 Answers 2

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The substitution $x=e^t$ is typically used when solving Cauchy-Euler equations.

If $x=e^t$ then $dx=e^tdt$ so $\dfrac{dt}{dx}=e^{-t}$.

Using the chain rule,

$$\dfrac{dy}{dx}=\dfrac{dt}{dx}\cdot\dfrac{dy}{dt}=e^{-t}\frac{dy}{dt}\tag{1}$$

Then for the second derivative we have

\begin{eqnarray} \frac{d^2y}{dx^2}&=&\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ &=&e^{-t}\left(\frac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\right) \text{ using equation } (1)\\ &=&e^{-t}\left(e^{-t}\frac{d^2y}{dt^2}-e^{-t}\frac{dy}{dt}\right)\\ &=&e^{-2t}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right) \end{eqnarray}

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If $x=e^{t}$, then by the chain rule we have

$$\frac{d}{dx}\{\cdot\}=\frac{1}{\frac{dx}{dt}}\frac{d}{dt}\{\cdot\} =e^{-t}\frac{d}{dt}\{\cdot\}$$

For the second derivative, we have

$$\begin{align} \frac{d^2}{dx^2}\{\cdot\}&=\frac{d}{dx}\left(\frac{d}{dx}\{\cdot\}\right)\\\\ &=\frac{d}{dx}\left(e^{-t}\frac{d}{dt}\{\cdot\}\right)\\\\ &e^{-t}\frac{d}{dt}\left(e^{-t}\frac{d}{dt}\{\cdot\}\right)\\\\ &=e^{-2t}\frac{d^2}{dt^2}\{\cdot\}-e^{-2t}\frac{d}{dt}\{\cdot\} \end{align}$$

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    Yes, I have the first one now. Its mainly the second derivative that I'm unsure of. What happens after d/dx(d/dx)? – 2017-02-10
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    @M.Turner Sure. We have successive applications of the operator in the first case. I've edited to show the steps. – 2017-02-10
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    Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – 2017-02-15