2nd derivative with total differential?

Question

Let the function $φ(x, y) = c$ be given. Now you have to determine the first and second derivative $y'$ and $y''$.

The first derivative worked out with the total differential: $φ_x \cdot dx + φ_y \cdot dy = 0\\ y' = dy/dx = - φ_x/φ_y$

Does anyone know how to determine y'' from this?

Answer 1

Assuming $x,y \in \mathbb{R}$, we have $\varphi:\mathbb{R}^2\to \mathbb{R}$. Now from $\varphi(x,y) = c$ we have $\frac{\partial \varphi}{\partial x} = \frac{\partial \varphi}{\partial xy} = 0$ and from there, nothing can be inferred from $x$ or $y$. So I thnik the problem is not posed correctly. I think the correct way to write what you are looking for is:

Let $\varphi:\mathbb{R}^2\to \mathbb{R}$ be a function and $y:\mathbb{R}\to \mathbb{R}$ a curve such that $\varphi(x,y(x))$ is constant. Find $y''(x)$.

Now this can be solved as follows:

For convenience we will use the notation: $\partial_x \varphi = \frac{\partial \varphi}{\partial x} = \varphi_x$ and $\partial_y \varphi = \frac{\partial \varphi}{\partial y} = \varphi_y$

Write $\psi(x) = \varphi(x,y(x)) = c$. Now $\psi'(x) = \partial_x\varphi(x,y(x)) + \partial_y \varphi(x,y(x)).y'(x) = 0$ (by the chain rule). This yields (for $\partial_y\varphi(x,y(x)) \neq 0$) $$y'(x) = - \frac{\partial_y\varphi(x,y(x))}{\partial_x\varphi(x,y(x))} = -\frac{\varphi_x(x,y(x))}{\varphi_y(x,y(x))}$$

And now we compute $y''(x)$ with the quotient rule and then the chain rule again:

$$y''(x) = \frac{d}{dx} \left(\frac{\varphi_x(x,y(x))}{\varphi_y(x,y(x))}\right)$$

(From now on lets drop the $y(x)$ just write $y$, but keep in mind it is still a function of $x$)

$$y''(x) =\frac{ \frac{d}{dx}(\varphi_x(x,y)).\varphi_y(x,y) - \frac{d}{dx}(\varphi_y(x,y)).\varphi_x(x,y)}{\varphi_y(x,y)^2}$$

$$y''(x) =\frac{1}{\varphi_y(x,y)^2} \left((\varphi_{xx}(x,y) + \varphi_{xy}(x,y).y').\varphi_y(x,y) - (\varphi_{yx}(x,y) + \varphi_{yy}(x,y).y').\varphi_x(x,y)\right)$$

Finally, replacing $y'$ by the expression previously obtained:

$$y''(x) =\frac{\left(\varphi_{xx}(x,y) - \varphi_{xy}(x,y)\frac{\varphi_x(x,y)}{\varphi_y(x,y)}\right)\varphi_y(x,y) - \left(\varphi_{yx}(x,y) - \varphi_{yy}(x,y)\frac{\varphi_x(x,y)}{\varphi_y(x,y)}\right)\varphi_x(x,y)}{\varphi_y(x,y)^2} $$ provided $\varphi$ is sufficiently differentiable.

Answer 2

You found $\dfrac{\mathrm d\varphi(x,y)}{\mathrm d x}=\varphi_x(x,y)+\varphi_y(x,y)\dfrac{\mathrm dy}{\mathrm dx}$ by the Partial Derivative Chain Rule, and since the left hand side was zero, that gave you:$$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\varphi_x(x,y)}{\varphi_y(x,y)}$$

This is a quotient of functions, so just apply the Derivative Quotient Rule:

$$\dfrac{\mathrm d^2 y}{\mathrm dx~^2} = - \left(\dfrac{\dfrac{\mathrm d \varphi_x(x,y)}{\mathrm d x}\cdot \varphi_y(x,y)-\varphi_x(x,y)\cdot\dfrac{\mathrm d \varphi_y(x,y)}{\mathrm d x}}{(\varphi_y(x,y))^2}\right)$$

Finish off by applying the Partial Derivative Chain Rule to those differential terms in the numerator.