## Vague Partial Derivative

Could someone please explain to me how to find the derivative of this:

dy/dx = φ(x, y)

Should I break up the equation to make it dy/dx = φ(x) + φ(y) and then derive the parts?

I would then get d²y/dx² = ∂φ/∂x + ∂φ/∂y
do I have to also multiply both terms by their respective derivatives of the inside variable?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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 Quote by lifhgrl823 dy/dx = φ(x, y) Should I break up the equation to make it dy/dx = φ(x) + φ(y) and then derive the parts?
If φ(x, y) is arbitrary why do you think you can break it up to φ(x) + φ(y)? If φ(x, y) = xy, how can this be broken up into φ(x) + φ(y)?
 That's a good point. My professor wrote that the second derivative should be: ∂φ/∂x + ∂φ/∂y (dy/dx) = ∂φ/∂x + φ(∂φ/∂x) I've been trying to play around with the equation and see how I could get that answer. All of the partial derivatives I've done previously had equations that were equal to f(x,y) or such.

Recognitions:
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## Vague Partial Derivative

 Quote by lifhgrl823 That's a good point. My professor wrote that the second derivative should be: ∂φ/∂x + ∂φ/∂y (dy/dx) = ∂φ/∂x + φ(∂φ/∂x) I've been trying to play around with the equation and see how I could get that answer. All of the partial derivatives I've done previously had equations that were equal to f(x,y) or such.
Can you see how the Professor gets the left side? It's the chain rule.
 Yes, to take the second derivative of y, you should look at it as phi(x,y(x)) so partial in x with respect to first entry, plus that with respect to second entry, which requires the chain rule.

 Tags calculus, math, multivariable, partial derivative