- #1

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I don't understand this as isn't according to chain rule,

So where is the

Source: http://www.math.ubc.ca/~feldman/m226/taylor2d.pdf

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- #1

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I don't understand this as isn't according to chain rule,

So where is the

Source: http://www.math.ubc.ca/~feldman/m226/taylor2d.pdf

- #2

blue_leaf77

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- #3

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Why is it not a function of t? I am still new to this so do you have something i can read on about this?

Besides thanks for your response!

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blue_leaf77

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It's not anWhy is it not a function of t?

$$

\frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} .

$$

Both answers should be identical.

I guess this problem should belong to multivariate calculus.do you have something i can read on about this?

- #5

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Thanks a lot! I now understand.It's not anexplicitfunction of ##t##. When you want to do chain rule, you have to pay attention on which variables are written explicitly, despite whether or not these variables are functions of yet another variable. For example take ##F(t) = xy^2## where ##x = \sqrt{t}## and ##y=t-2##. If you want to calculate ##dF/dt##, you can either first express ##x## and ##y## in terms of ##t## and then differentiate w.r.t. ##t## or let ##F## be expressed in ##x## and ##y## then use the chain rule

$$

\frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} .

$$

Both answers should be identical.

I guess this problem should belong to multivariate calculus.

- #6

Svein

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I can agree with the second line, but the third line is pure nonsense.

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blue_leaf77

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In the linked pdf file, the author for some reason sets ##\Delta t = 1##.

I can agree with the second line, but the third line is pure nonsense.

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