Chain rule

  • Thread starter Joseph1739
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(df/dx) + (df/dy)* (dy/dx) = df(x,y)/dx

My book mentions the chain rule to obtain the right side of the equation, but I don't see how. The chain rule has no mention of addition. The furthest I got was applying the chain rule to the right operant resulting in:

df/dx + df/dx = 2(df/dx)
 

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  • #2
Samy_A
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(df/dx) + (df/dy)* (dy/dx) = df(x,y)/dx

My book mentions the chain rule to obtain the right side of the equation, but I don't see how. The chain rule has no mention of addition. The furthest I got was applying the chain rule to the right operant resulting in:

df/dx + df/dx = 2(df/dx)
I'm trying to make sense of this equation.

The only thing I can think of is this:
##f## is a function of two variables, ##x## and ##y##, and then they consider that ##y## is some function of ##x##.
What they actually compute is the derivative of ##g(x)=f(x,y(x))##.
For that, the chain rule can be used, giving:
##\frac{dg}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \frac{dy}{dx} ##.

It's a little confusing, as ##y## is both the name of the second variable and considered a function of ##x##. And they don't use a new name ##g##, but keep ##f##.
That's how they get ##\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \frac{dy}{dx} ##.

It's somewhat sloppy, as they don't use ##\partial## for what clearly must be partial derivatives.
 

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