integral rules

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JeckShirauw

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Could someone please explain to me why the following is true?
y^2=x
2y dy/dx = 1

I do not understand why the y^2 becomes 2y

Thank you very much in advance.
 
What do you get for this? [math]\dfrac{d}{dx} x^{2} = [/math]
 
No worries. This is why "Implicit Differentiation" was invented!

If two things are equal, and their derivatives exist, why wouldn't their derivatives be equal?

"Implicit Differentiation" has a natural with the "Chain Rule".

Given [math]y^{2} = x[/math], our task is to find dy/dx. To do this, we simply put in our pocket the assumption that y = f(x), or that y is some function of x, and we can proceed. This is one of the most beautiful things about mathematics, in my opinion. Simply assuming that we know something leads to a solution and we are rewarded for our boldness by determining what the something actually is! It's just totally cool that this would work!!

Given that y = f(x), the chain rule tells us, [math]\dfrac{d}{dx}y^{2} = 2y\cdot\dfrac{dy}{dx}[/math]. That last dy/dx is our assumption that f'(x) exists and we are calling it dy/dx.

Notice how I haven't even mentioned the other side of the equation. I'm assuming we know that dx/dx = 1.
 
It means "the derivative of y with respect to x" where y is a function of the independent variable x! That's probably what you learned. If you are looking for something deeper it is "the rate of change of y relative to x". How fast y changes as x changes. For example, if y is a constant, the it doesn't change so \(\displaystyle \frac{dy}{dx}= 0\). If y= x, y changes exactly as fast as x so \(\displaystyle \frac{dy}{dx}= 1\). If y= 5x then y changes 5 times as fast as x (if x changes from 1 to 2, y changes from 5(1)= 5 to 5(2)= 10 so changes by 5, 5 times as much as x) so \(\displaystyle \frac{dy}{dx}= 5\).

For linear functions, all that is very easy and can be done with algebra. For non-linear functions it is not so easy. For example, if \(\displaystyle y= x^2\) and x changes from 1 to 2, y changes from 1 to 4 so has changed by 4- 1= 3 while x changed by 1. y changed by 3 times as much as x. But if x changes from 2 to 3, y changes from 4 to 9. y has changed by 5 while x still changed by 1. Now, y has changed by 5 times as much as x. The "rate of change of y compared to x" is, for y a non-linear function, a function of x rather than a constant. That is the situation where we need "Calculus".

pka, I am not shocked by the question. I have known many students who learned the "mechanics" of Calculus without actually learning the "meaning". Jeckshirauw is, at least asking that question!
 
Let me add a little more. Isaac Newton, and many others of his time, was interested in the movements of the planets. Kepler and Copernicus has done detailed studies of the actual motions, giving empirical calculations of the motions. Newton wanted to determine a law that would explain those motions. He came up with his famous formula, \(\displaystyle F=-\frac{GmM}{r^2}\).

But there is a complication hidden there. It says that we can calculate the force for specific "r", the distance of the planet from the sun. But, of course, the planet is moving! "r" is only correct for one instant. That formula gives the force at a given instant. But F= ma. A given force, f, gives a specific acceleration, a. Acceleration is the rate of change of velocity and velocity is the rate of change of position. How can you have an acceleration, so a velocity at one instant? Velocity is (or was before Newton) the distance you moved divided by the time it took you to move that distance. That necessarily requires a change in time. With that definition it would be impossible to talk about velocity, or acceleration, at a given instant. Calculus allows us to talk about (and hopefully understand what we mean by) the instantaneous velocity and acceleration of an object.
 
You always (OK, usually) have x and y in the equation. Remember y = f(x), that is whenever an equation has f(x) it has y. So if you have f(x) = 3x+7, then y = 3x+7. The derivative of y wrt x is y' or dy/dx. The derivative of 3x+7 wrt x is 3dx/dx or 3. So in the end you have y'=3.
 
Yes, students in calculus can get decent grades and not know what a derivative at any time during the course. They do so by memorizing the entire course. It is sad that this possible but it happens all the time. I hated giving out A-'s and B's to students who knew very little Calculus but could do well on exams.
 
Thank you very much!

The only thing I didn't understand quite well is
"The derivative of 3x+7 wrt x is 3dx/dx or 3. "

dx/dx means the derivative of x with respect to x, right?
So what I think is that if x changes with 1, then x changes with 1 and not 3.
3x+7 does change with 3, but then I need to write d(3x+7)/dx and not dx/dx?

Can you please explain what I misunderstood?

Sorry if it is a weird question.
 
Let's build a Newton quotient

[MATH]\{3(x + h) + 7\} - (3x + 7) = 3x + 3h + 7 - (3x + 7) = 3h + 3x - 3x + 7 - 7 = 3h.[/MATH]
Make sense?

[MATH]\dfrac{3h}{h} = 3.[/MATH]
OK with that?

[MATH]\lim_{h \rightarrow 0} 3 = 3.[/MATH]
Three stays three no matter what h does.

[MATH]\therefore \dfrac{d(3x + 7)}{dx} = 3.[/MATH]
It is as simple as that. The derivative is the limit of the Newton quotient as h approaches 0.
 
JeckShirauw said:
Sorry if it is a weird question.
Click to expand...
It's not "weird" but there are a lot of misuns3erstandings!

JeckShirauw said:
dx/dx means the derivative of x with respect to x, right?
So what I think is that if x changes with 1, then x changes with 1 and not 3.
Click to expand...
Yes, dx/dx=1 but what you have above is "3dx/dx" which is 3 times dx/dx. 3 times 1 is 3.

3x+7 does change with 3, but then I need to write d(3x+7)/dx and not dx/dx?
Click to expand...
I am not clear what your question is. And "change with 3" is not standard English! If x "changes by 1" then x changes, say, from x= 3 to x= 4, then 3x+ 7 will change from 3*3+ 7= 16 to 3*4+ 7= 19 so has changed by 19- 16= 3.

Can you please explain what I misunderstood?
Click to expand...
I am a bit surprised by your reference to d(3x+ 7)/dx. Usually before you learn derivatives, you learn about the "slope" of a straight line. The slope of a straight line, y= ax+ b, is the tangent of the angle it makes with the x-axis. Of course, the basic definition of tangent of an angle in a right triangle is "opposite side over near side". On a graph, the "opposite side" is the "rise", the increase in y. and the "near side" is the "run", the increase in x, so that the slope is "change in y divided by change in x". If we increase x by h, from x to x+ h, then y= ax+ b changes from ax+ b to a(x+ h)+ b= ax+ ah+ b. The change in y is ax+ ah+b- ax- b= ah while the change in x is h. The slope is ah/h= a.

But that doesn't require any Calculus or any dy/dx! Calculus comes into play when we have non-linear functions.
 
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