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- Thread starter Bashyboy
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[tex]\frac{d}{dx}f(y)[/tex]

Which after the chain rule becomes,

[tex]\frac{df(y)}{dx} = \frac{df(y)}{dy} \cdot \frac{dy}{dx}[/tex]

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For instance, [itex]\frac{d}{dy}[y^2][/itex] Which would equal [itex]2y\frac{dy}{dx}[/itex] Where the 2y is how y changes with respect to itself, and the other part is how y changes with x?

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An excerpt from my calculus books reads, "...when you differentiate terms involving y, you must first apply the chain rule, because you are assuming that y is defined implicitly as a differentiable function of x." Why are we allowed to assume in this case? And what does it mean to assume this?

EDIT:

I just read on a website that a function that is implicit not a function that is not depend on x, but it's just not exactly clear how it is dependent upon x. Is that why we can assume, because even though it is not clear how y is depending on x, it still is, we just can't show exactly how it depends on x?

EDIT:

I just read on a website that a function that is implicit not a function that is not depend on x, but it's just not exactly clear how it is dependent upon x. Is that why we can assume, because even though it is not clear how y is depending on x, it still is, we just can't show exactly how it depends on x?

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that is equal to 2y because you differentiate wrt y.For instance, [itex]\frac{d}{dy}[y^2][/itex] Which would equal [itex]2y\frac{dy}{dx}[/itex] Where the 2y is how y changes with respect to itself, and the other part is how y changes with x?

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HallsofIvy

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What you have written is not correct, but may just be a typo. You meant to write [itex]\frac{d}{dx}[y^2][/itex], not [itex]\frac{d}{dy}[y^2][/itex].

For instance, [itex]\frac{d}{dy}[y^2][/itex] Which would equal [itex]2y\frac{dy}{dx}[/itex] Where the 2y is how y changes with respect to itself, and the other part is how y changes with x?

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AlephZero

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If you write an equation that shows "exactly" how y depends on x, that doesn't mean you can turn it into an equation that says "y = a formula that only contains x".

Try it yourself with an equation like ##\sin(x/y) + \cos(y/x) = 0## for example.

An "implicit function" is where the x's and y's mixed up together. An "explicit function" has the x's and y's separated out from each other.

Try it yourself with an equation like ##\sin(x/y) + \cos(y/x) = 0## for example.

An "implicit function" is where the x's and y's mixed up together. An "explicit function" has the x's and y's separated out from each other.

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What you have written is not correct, but may just be a typo. You meant to write [itex]\frac{d}{dx}[y^2][/itex], not [itex]\frac{d}{dy}[y^2][/itex].

HallsofIvy, could you possible explain the procedure of the example problem given in the link?

http://www.mash.dept.shef.ac.uk/Resources/web-implicit.pdf

Page 4 section 3.

In the example problem, I know they ultimately want to ascertain how the function changes with x--or, in other words, take the derivative with respect to x. But you can see that the first take the derivative with respect to y for the y terms and then multiply by dy/dx. Actually, in their steps to implicit differentiation, they state that, "to differentiate a function of y with respect to x, we differentiate with respect to y and then multiply by dy/dx. I don't understand the point in differentiating with respect to y first, then multiplying by dy/dx.

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The derivative of y

Then they rearranged the equation so that the y' are on one side. They then factor out a y' and divide by the stuff that's left.

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I don't understand the point in differentiating with respect to y first, then multiplying by dy/dx.

As said before, it is because of the chain rule.

When you have a function which depends on y, f = f(y), where y depends on x, y = y(x), the derivative df/dx can be multiplied by dy/dy (which is 1 and so does not change the value) and reorganized:

[tex]\frac{df}{dx} = \frac{df}{dx}\frac{dy}{dy} = \frac{df}{dy}\frac{dy}{dx}[/tex]

This is used in the example to find the dependance that y has on x, given an equation of only y, x and constants.

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HallsofIvy

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Oh, yes, very interesting. Thank you everyone.

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