How to solve this second derivative?

Question

I want to solve the derivative $\frac{d^{2}\psi}{dx^{2}}$ for $\psi(x) = Ae^{iu(x)}$. Here, A is a constant, $u = u(x)$ and $i$ is the complex number.

The answer for this has been given as: $i \frac{d^{2}u(x)}{dx^{2}} \psi(x) - (\frac{du(x)}{dx})^{2}\psi(x)$.

However, I keep getting something like this:

$$2i\frac{du(x)}{dx}(\frac{d^{2}u(x)}{dx^{2}})\psi(x) - (\frac{du(x)}{dx})^{4}\psi(x).$$

I used the chain rule and then the product rule to get this.

Could someone start me off on the right path with this?

Answer 1

Using the chain rule, we have $$\frac{dψ}{dx} = Ai\left(\frac{du}{dx}\right)e^{iu(x)} = i\left(\frac{du}{dx}\right)ψ(x).$$ Then, using the product rule and the above, $$\frac{d^2ψ}{dx^2} = i\left(\frac{d^2u}{dx^2}\right)ψ(x) + i\left(\frac{du}{dx}\right)\frac{dψ}{dx}.$$ Now with $i^2 = -1$ we conclude $$\frac{d^2ψ}{dx^2} = i\left(\frac{d^2u}{dx^2}\right)ψ(x) + i\left(\frac{du}{dx}\right)\cdot i\left(\frac{du}{dx}\right)ψ(x) = i\left(\frac{d^2u}{dx^2}\right)ψ(x) - \left(\frac{du}{dx}\right)^2 ψ(x).$$

Answer 2

Chain rule can work but is somewhat complicated. Here's a mostly complete alternative approach. Plug in the lower derivatives of $\psi$ to get a more complete answer.

$\ln \psi =\ln A + iu$

$\frac{\psi '}{\psi}=iu'\implies \psi'=iu'\psi$

$\psi'' = iu''\psi+iu'\psi'$

Answer 3

\begin{align} \psi(x) &= Ae^{-iu(x)} \\ \psi'(x) &= Ae^{-iu(x)}(-iu'(x))=-Aiu'(x)e^{-iu(x)} \\ \psi''(x) &= -Ai\left(\frac{d}{dx}u'(x)\right)e^{-iu(x)}-Aiu'(x)\left(\frac{d}{dx}e^{-iu(x)}\right) \\ &=-Aiu''(x)e^{-iu(x)}-Aiu'(x)\left(-iu'(x)e^{-iu(x)}\right) \\ &= -Aiu''(x)e^{-u(x)}-A\big(u'(x)\big)^2e^{-iu(x)} \end{align}