Higher order Legendre's linear differential equation

Question

I understand that a Legendre's linear equation has the following form:

$a_n (\alpha x + \beta) \frac{d^n y}{dx^n} +....+ a_1(\alpha x + \beta)\frac{dy}{dx} + a_0 y = f(x) $

where $\alpha, \beta$ and $a_n$ are constant terms. In order to solve this kind of equation my textbook suggests the following substitution: $ \alpha x + \beta = e^t$

$\therefore \frac{dy}{dx} = \frac{dt}{dx} \frac{dy}{dt} = \frac{\alpha}{\alpha x + \beta} \frac{dy}{dt}$

And then it suggests looking at the higher order derivatives, for example for order 2:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} = (\frac{\alpha}{\alpha x + \beta})^2 (\frac{d^2y}{dt^2} - \frac{dy}{dt} )$

However how is this possible? Shouldn't it be:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} = (\frac{\alpha}{\alpha x + \beta})^2 \frac{dy}{dt} + \frac{d^2y}{dt dx} (\frac{\alpha}{\alpha x + \beta})$

Answer 1

\begin{align} \frac{d^2y}{dx^2}&=\frac{d}{dx} \frac{dy}{dx}\\ & = \frac{d}{dx}\Bigl(\frac{\alpha}{\alpha x + \beta}\,\frac{dy}{dt}\Bigr)\\ &=-\bigl(\frac{\alpha}{\alpha x + \beta}\Bigr)^2 \frac{dy}{dt} + \frac{\alpha}{\alpha x + \beta}\frac{d}{dx}\frac{dy}{dt}\\ &=-\bigl(\frac{\alpha}{\alpha x + \beta}\Bigr)^2 \frac{dy}{dt} +\bigl(\frac{\alpha}{\alpha x + \beta}\Bigr)^2 \frac{d^2 y}{dt^2}\ . \end{align}