Finding a second derivative using implicit differentiation with the chain rule

Question

I'm trying to find the second derivative $d^2y/dx^2$. In my problem, $y = y(x)$ and we are supposed to use the substitution of $\alpha x + \beta = e^t$ for some independent $t$.

(This pertains to solving Legendre's linear differential equation)

So, let $\alpha x + \beta = e^t$. Implicitly differentiating this with respect to $y$: $$\frac{d}{dy} (\alpha x + \beta) = \frac{d}{dy}(e^t)$$ $$\alpha \frac{dx}{dy} = e^t \frac{dt}{dy} = (\alpha x + \beta) \frac{dt}{dy}$$ Rearranging to express for $dy / dx$ $$\boxed{\frac{dy}{dx} = \frac{\alpha}{\alpha x + \beta} \frac{dy}{dt}}$$

Now, I need to find the $d^2 y / dx^2$. I can differentiate the LHS easy enough, but the RHS seems strange. I know that $x = x(t)$ and $y = y(x)$ but I'm unsure how to apply the chain rule here.

Answer 1

First apply the Product Rule.

$\def\d{\operatorname d}\qquad\begin{align}\dfrac{\d^2y}{\d x^2}&=\dfrac{\d~~}{\d x}\left[\dfrac{\alpha}{\alpha x+\beta}\cdot\dfrac{\d y}{\d t}\right]\\[1ex]&=\dfrac{\d~~}{\d x}\left[\dfrac{\alpha}{\alpha x+\beta}\right]\cdot\dfrac{\d y}{\d t}+\dfrac{\alpha}{\alpha x+\beta}\cdot\dfrac{\d~~}{\d x}\left[\dfrac{\d y}{\d t}\right]\end{align}$

Now apply the Chain Rule:

Answer 2

Another approach would start from the relation you produced, $$ \alpha \frac{dx}{dy} \ \ = \ \ e^t \frac{dt}{dy} \ \ = \ \ (\alpha x + \beta)· \frac{dt}{dy} $$ $$ \Rightarrow \ \ \frac{d}{dx} \ \left[ \ \frac{1}{\alpha}· \frac{dy}{dx} \ \right] \ \ = \ \ \frac{d}{dx} \ \left[ \ e^{-t} · \frac{dy}{dt} \ \right] \ \ = \ \ \frac{dt}{dx} · \frac{d}{dt} \ \left[ \ e^{-t} · \frac{dy}{dt} \ \right] $$ $$ = \ \ \frac{dt}{dx}· \left( \ -e^{-t} · \frac{dy}{dt} \ + \ e^{-t} · \frac{d^2y}{dt^2} \ \right) $$ $$ \Rightarrow \ \ \frac{1}{\alpha}· \frac{d^2y}{dx^2} \ \ = \ \ \frac{dt}{dx}·e^{-t} · \left( \ \frac{d^2y}{dt^2} \ - \ \frac{dy}{dt} \ \right) $$

[ $ \ \large{\frac{dt}{d x} \ = \ \frac{\alpha}{\alpha x \ + \ \beta} } \ \ , \ $ as noted in Graham Kemp's comment to the OP]

$$ \Rightarrow \ \ \frac{d^2y}{dx^2} \ \ = \ \ \alpha· \left(\frac{\alpha}{\alpha x \ + \ \beta} \ \right)·\left(\frac{1}{\alpha x \ + \ \beta} \ \right) · \left( \ \frac{d^2y}{dt^2} \ - \ \frac{dy}{dt} \ \right) $$ $$ = \ \ \left(\frac{\alpha}{\alpha x \ + \ \beta} \ \right)^2· \left( \ \frac{d^2y}{dt^2} \ - \ \frac{dy}{dt} \ \right) \ \ . $$

[This also permits us to write

$$ \frac{d^2y}{dx^2} \ + \ \left(\frac{\alpha}{\alpha x \ + \ \beta} \ \right)·\frac{dy}{dx} \ \ = \ \ \left(\frac{\alpha}{\alpha x \ + \ \beta} \ \right)^2 · \frac{d^2y}{dt^2} \ \ . \ ]$$

This is in agreement with Graham Kemp's relation, $$ \frac{d^2y}{dx^2} \ \ = \ \ \frac{d}{d x}\left[\frac{\alpha}{\alpha x + \beta}\right]\cdot\frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{d}{d x}\left[\frac{d y}{d t} \right]$$ $$ = \ \ \alpha·\left[ \ -(\alpha x + \beta)^{-2}·\alpha \ \right]\cdot\frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{dt}{d x}\cdot\frac{d}{d t}\left[\frac{d y}{d t} \right]$$ $$ = \ \ -\left(\frac{\alpha}{\alpha x + \beta} \right)^2 \cdot \frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{d^2 y}{d t^2} \ \ . $$