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 November 14th, 2017, 02:10 AM #11 Senior Member   Joined: Nov 2011 Posts: 110 Thanks: 1 Rock on What is "rock on"?
 November 14th, 2017, 02:16 AM #12 Senior Member   Joined: Nov 2011 Posts: 110 Thanks: 1 There are 4 definitions that connect one to another: 1. To solve the equation you need to seprate the y. 2. It a function in a range of value that strange from the "normal" function by the range that it defines. It a function with a limit of range of value. 3. It is connected to the Implicit functions Theorem. 4. It has 2 inputs (x & y) and one output [I think it a y but it can be also z]. All the definition [4 in particular] in two "outputs": x and y. The noition of it is F(x, y) = 0. Understoond?
November 14th, 2017, 02:37 PM   #13
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Quote:
 Originally Posted by shaharhada What is "rock on"?
It means we are in agreement and that Physics is awesome. So "rock on" means "keep on doing it."

-Dan

November 15th, 2017, 05:06 AM   #14
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If the "strange font" was $\displaystyle \partial$, as in $\displaystyle \frac{\partial f}{\partial x}$, that is, in fact, the partial derivative. Here, f is a function of more than a single variable (say, x and y) and $\displaystyle \frac{\partial f}{\partial x}$ is the "rate of change of f as x changes but y does not change". That is, we treat y as a constant rather than a variable. Similarly we find $\displaystyle \frac{\partial f}{\partial y}$ by treating x as a variable and thinking of f as a function of y only.

For example, if $\displaystyle f(x,y)= x^3y^2$, since the derivative of $\displaystyle Cx^3$ is $\displaystyle 3Cx^2$ for any constant C, the "partial derivative with respect to x" is $\displaystyle \frac{\partial f}{\partial x}= 3x^2y^2$. Similarly, since the derivative of $\displaystyle Cy^2$ is $\displaystyle 2Cy$ for any constant C, $\displaystyle \frac{\partial f}{\partial y}= 2x^3y$.

Quote:
 Originally Posted by shaharhada There are 4 definitions that connect one to another: 1. To solve the equation you need to seprate the y.
What equation are you talking about? You said nothing about "solving an equation" before.

Quote:
 2. It a function in a range of value that strange from the "normal" function by the range that it defines. It a function with a limit of range of value.
I don't understand what you mean by this. If you are talking about a function of two or more variables, it is the domain that is "strange" ($\displaystyle R^2$ or $\displaystyle R^3$ rather than $\displaystyle R$). The range is usually just $\displaystyle R$ as in the one variable case.

Quote:
 3. It is connected to the Implicit functions Theorem.
"Connected" to what? It is true that you can use the "Implicit Function Theorem" to show that, under certain conditions, f(x, y)= C can be solved for x or y.

Quote:
 4. It has 2 inputs (x & y) and one output [I think it a y but it can be also z].
If you are using "y" to mean one of the inputs, you would certainly NOT use y again to represent the output!

Quote:
 All the definition [4 in particular] in two "outputs": x and y. The noition of it is F(x, y) = 0. Understoond?
You just said, in 4, that x and y were inputs, not "outputs"!

If you have a function of two variables (two inputs), F(x, y), set equal to a constant, F(x, y)= C (possibly 0 but not necessarily) then the "implicit function theorem" says that "if $\displaystyle \frac{\partial f}{\partial y}$ is not 0 in some region around $\displaystyle (x_0, y_0)$ then we can find some function of a single variable, f, such that y= f(x) in that region. That is, F(x, f(x))= C".

 November 16th, 2017, 02:38 AM #15 Senior Member   Joined: Nov 2011 Posts: 110 Thanks: 1 Thanks for the maintance. You are welcome!!!! In 2 I still think that the functions is limited. Show me Implicit Function that doesn't limited. In the end of the post it is a writing error that "inputs" become "outputs". Last edited by shaharhada; November 16th, 2017 at 02:41 AM.
 November 16th, 2017, 04:59 AM #16 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,821 Thanks: 750 "Limited"? Do you mean "bounded"? The functions defined implicitly by x+ y= 0 are unbounded.
November 16th, 2017, 05:07 AM   #17
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Quote:
 Originally Posted by Country Boy "Limited"? Do you mean "bounded"? The functions defined implicitly by x+ y= 0 are unbounded.
x + y = 0 is not implicit function.
I think you are wrong.
It can be written as y = -x.
It is a regular function, not?

 November 16th, 2017, 05:21 AM #18 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,821 Thanks: 750 Then what do you mean by "implicit function"? Personally, I would not use the phrase "implicit function" but rather "implicitly defined" function. A function is "implicitly defined" if it is given as f(x, y)= C which can be solved for y as a function of x. Perhaps you want an example where there exist more than one way to define y as a function of x? For example, $f(x,y)= x^2+ y^2= 1$ where y can be either $y= \sqrt{1- x^2}$ or $y= -\sqrt{1- x^2}$.
 November 16th, 2017, 05:49 AM #19 Senior Member   Joined: Nov 2011 Posts: 110 Thanks: 1 Yes!!!

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