Second derivative, chain rules and order of operations

In summary, the chain rule for second derivatives states that the derivative of a function with respect to another function is the sum of the derivatives of the two functions with respect to the same variable.
  • #1
Fady Megally
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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 ?
 
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  • #2
Fady Megally said:
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.
 
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  • #3
anuttarasammyak said:
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
 
  • #4
Now I can find y there. thanks.
Say y=y(x) and x=x(t)
[tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}...(1)[/tex] and so
[tex]\frac{d^2y}{dt^2}= \frac{d}{dt}[\frac{dy}{dx}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}...(2)[/tex]
[tex]= \frac{d}{dx}[\frac{dy}{dx}\frac{dx}{dt}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}[/tex]
[tex]= \frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}[/tex]
as you wrote. From (1) replacing y with C
[tex]\frac{dC}{dt}=\frac{dC}{dx}\frac{dx}{dt}[/tex]
and (2) is
[tex]\frac{d^2C}{dt^2}= \frac{d}{dt}[\frac{dC}{dx}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}[/tex]
[tex]= \frac{d}{dx}[\frac{dC}{dt}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}[/tex]
changing order of application of d/dx and d/dt to C in the first RHS term. Now we could follow the text.
 
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  • #5
anuttarasammyak said:
Now I can find y there. thanks.
Say y=y(x) and x=x(t)
[tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}...(1)[/tex] and so
[tex]\frac{d^2y}{dt^2}= \frac{d}{dt}[\frac{dy}{dx}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}...(2)[/tex]
[tex]= \frac{d}{dx}[\frac{dy}{dx}\frac{dx}{dt}]\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}[/tex]
[tex]= \frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}[/tex]
as you wrote. From (1) replacing y with C
[tex]\frac{dC}{dt}=\frac{dC}{dx}\frac{dx}{dt}[/tex]
and (2) is
[tex]\frac{d^2C}{dt^2}= \frac{d}{dt}[\frac{dC}{dx}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}[/tex]
[tex]= \frac{d}{dx}[\frac{dC}{dt}]\frac{dx}{dt}+\frac{dC}{dx}\frac{d^2x}{dt^2}[/tex]
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
$$\frac {d(\frac {dy} {dx})}{dx} \cdot \frac {dx} {dt} =\frac {d(\frac {dy} {dx}\cdot \frac {dx} {dt})}{dx}$$

??
 
  • #6
Well, in simple
[tex]\frac{d}{dx}[\frac{dy}{dt}]=\frac{d}{dt}[\frac{dy}{dx}][/tex]
 
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  • #7
Fady Megally said:
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 [tex]
\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}[/tex] 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 [tex]
\dot{\frac{dC}{dx}} = \frac{d}{dt}\frac{dC}{dx} = \dot x \frac{d^2C}{dx^2}.[/tex] Note the difference in the position of the dot: in the first it is over the [itex]C[/itex], and in the second it is over the entire [itex]\frac{dC}{dx}[/itex].
 

FAQ: Second derivative, chain rules and order of operations

1. What is a second derivative?

A second derivative is the derivative of a derivative. It measures the rate of change of the rate of change of a function. In other words, it tells us how fast the slope of a function is changing at a specific point.

2. How do you find the second derivative of a function?

To find the second derivative of a function, you first need to find the first derivative using the power rule, product rule, or quotient rule. Then, you can find the second derivative by taking the derivative of the first derivative using the same rules. Alternatively, you can use the notation f''(x) to represent the second derivative of a function f(x).

3. What is the chain rule?

The chain rule is a rule used to find the derivative of a composite function, which is a function that is made up of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

4. How do you apply the chain rule to find the second derivative?

To apply the chain rule to find the second derivative, you first need to find the first derivative using the chain rule. Then, you can use the chain rule again to find the second derivative. This process can be repeated to find higher order derivatives.

5. What is the order of operations for finding derivatives?

The order of operations for finding derivatives is as follows: first, simplify the function if possible; then, apply any applicable rules such as the power rule, product rule, quotient rule, or chain rule; and finally, evaluate the derivative at a specific point if needed. It is important to follow this order to ensure accurate results.

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