Second derivative, chain rules and order of operations

[Fady Megally]

So the chain rule for second derivatives is $$ \frac {d^2 y} {d t^2} = \frac{d}{dx}(\frac {dy} {dx}) \cdot \frac {dx} {dt} \cdot \frac {dx} {dt} + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2} = \frac{d^2 y}{d x^2} \cdot (\frac {dx} {dt})^2 + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2}$$

Today I came across this equation in a graphics/computer modeling course
$$\ddot C = \frac {d\dot C} {dx} \cdot \dot x + \frac {dC} {dx} \cdot \ddot x$$

Now what i would infer from this is that
$$\frac {d} {dx}(\frac {dy} {dx}) \cdot \frac {dx} {dt} = \frac {d\dot y} { dx}$$

This sounds right but can someone point me to a rule or theorem that suggests that. or a proof maybe ?

 

[anuttarasammyak]

So the chain rule for second derivatives is

The formula you typed has something wrong at least at the first term of RHS ( no y in it). Typo ? Please check it out.

 

[Fady Megally]

The formula you typed has something wrong at least at the first term of RHS ( no y in it). Typo ? Please check it out.

I don't think so .. what i mean by $$\frac{d}{dx}(\frac {dy} {dx})$$ is $$\frac {d (\frac {dy} {dx})} {dx}$$

i think this is what you were pointing to ? sorry if i missed something else

 

[anuttarasammyak]

Now I can find y there. thanks.
Say y=y(x) and x=x(t)
$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}...(1)$$ and so
$$\frac{d^2y}{dt^2}= \frac{d}{dt}[\frac{dy}{dx}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}...(2)$$
$$= \frac{d}{dx}[\frac{dy}{dx}\frac{dx}{dt}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}$$
$$= \frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}$$
as you wrote. From (1) replacing y with C
$$\frac{dC}{dt}=\frac{dC}{dx}\frac{dx}{dt}$$
and (2) is
$$\frac{d^2C}{dt^2}= \frac{d}{dt}[\frac{dC}{dx}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}$$
$$= \frac{d}{dx}[\frac{dC}{dt}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}$$
changing order of application of d/dx and d/dt to C in the first RHS term. Now we could follow the text.

 

[Fady Megally]

Now I can find y there. thanks.
Say y=y(x) and x=x(t)
$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}...(1)$$ and so
$$\frac{d^2y}{dt^2}= \frac{d}{dt}[\frac{dy}{dx}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}...(2)$$
$$= \frac{d}{dx}[\frac{dy}{dx}\frac{dx}{dt}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}$$
$$= \frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}$$
as you wrote. From (1) replacing y with C
$$\frac{dC}{dt}=\frac{dC}{dx}\frac{dx}{dt}$$
and (2) is
$$\frac{d^2C}{dt^2}= \frac{d}{dt}[\frac{dC}{dx}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}$$
$$= \frac{d}{dx}[\frac{dC}{dt}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}$$
changing order of application of d/dx and d/dt to C in the first RHS term. Now we could follow the text.

Thanks anuttarasammyak

So what you are saying that moving the dx/dt in and out of the d/dx() operator is valid ?? that's to say
$$\frac {d(\frac {dy} {dx})}{dx} \cdot \frac {dx} {dt} =\frac {d(\frac {dy} {dx}\cdot \frac {dx} {dt})}{dx}$$

??

 

[anuttarasammyak]

Well, in simple
$$\frac{d}{dx}[\frac{dy}{dt}]=\frac{d}{dt}[\frac{dy}{dx}]$$

 

[pasmith]

So the chain rule for second derivatives is $$ \frac {d^2 y} {d t^2} = \frac{d}{dx}(\frac {dy} {dx}) \cdot \frac {dx} {dt} \cdot \frac {dx} {dt} + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2} = \frac{d^2 y}{d x^2} \cdot (\frac {dx} {dt})^2 + \frac {dy} {dx} \cdot \frac {d^2 x} {d t^2}$$

Today I came across this equation in a graphics/computer modeling course
$$\ddot C = \frac {d\dot C} {dx} \cdot \dot x + \frac {dC} {dx} \cdot \ddot x$$

I would interpret that as $$\frac{d\dot C}{dx} = \frac{d}{dx}\frac{dC}{dt} = \frac{d}{dx} \left(\dot x\frac{dC}{dx}\right) = \dot x\frac{d^2C}{dx^2} + \frac{dC}{dx}\frac{\ddot x}{\dot x}$$ which does not lead to a correct statement of the chain rule, whereas I'm sure what the author meant (and possibly actually wrote) is $$\dot{\frac{dC}{dx}} = \frac{d}{dt}\frac{dC}{dx} = \dot x \frac{d^2C}{dx^2}.$$ Note the difference in the position of the dot: in the first it is over the $C$, and in the second it is over the entire $\frac{dC}{dx}$.