- #1

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y'' = partial f / partial x + partial f / partial y * y'

But at an intuitive level I cannnot understand why. How does the second derivative equal the sum of the two partial derivatives times the first?

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- Thread starter bangthatdrum
- Start date

- #1

- 11

- 0

y'' = partial f / partial x + partial f / partial y * y'

But at an intuitive level I cannnot understand why. How does the second derivative equal the sum of the two partial derivatives times the first?

- #2

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Please ignore this.

- #3

- 15,393

- 686

Numerical initial value techniques such as Runge-Kutta address the problem of approximating

[tex]\aligned

y'(t) &\equiv \frac{dy(t)}{dt} = f(t,y(t)) \\[4pt]

y(t_0) &= y_0

\endaligned[/tex]

Suppose some function

[tex]\frac{dg(t,y(t))}{dt} = \frac{\partial g(t,y)}{\partial t} + \frac{\partial g(t,y)}{\partial y}\frac{dy}{dt}[/tex]

With this, the second time derivative of

[tex]y''(t) = \frac{dy'}{dt} = \frac{df(t,y(t))}{dt} = \frac{\partial f(t,y)}{\partial t} + \frac{\partial f(t,y)}{\partial y}\frac{dy}{dt}[/tex]

But we already know

[tex]y''(t) = \frac{\partial f(t,y)}{\partial t} + \frac{\partial f(t,y)}{\partial y}f(t,y)[/tex]

- #4

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- #5

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Thank you D H. This is wonderful.

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