# Finding a second derivative using implicit differentiation with the chain rule

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.

• What is the input, what functional dependency is known at the start? You say $y(x)$, but in the computation it looks like you start with the dependency $y(t)$. Commented Jul 16, 2022 at 14:37
• Apply the product rule before the chain rule. Commented Jul 17, 2022 at 7:55
• Also, recognise that you have already found $\dfrac{\mathrm d t}{\mathrm d x}=\dfrac{\alpha}{\alpha x+\beta}$. Commented Jul 17, 2022 at 8:16
• How are you going with this? Commented Jul 19, 2022 at 4:52
• @nocomment Have you made any progress? Commented Jul 20, 2022 at 1:30

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:

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, $$\frac{d^2y}{dx^2} \ \ = \ \ \frac{d}{d x}\left[\frac{\alpha}{\alpha x + \beta}\right]\cdot\frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{d}{d x}\left[\frac{d y}{d t} \right]$$ $$= \ \ \alpha·\left[ \ -(\alpha x + \beta)^{-2}·\alpha \ \right]\cdot\frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{dt}{d x}\cdot\frac{d}{d t}\left[\frac{d y}{d t} \right]$$ $$= \ \ -\left(\frac{\alpha}{\alpha x + \beta} \right)^2 \cdot \frac{d y}{d t} \ + \ \left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\left(\frac{\alpha}{\alpha x + \beta} \right)\cdot\frac{d^2 y}{d t^2} \ \ .$$