derivative meaning

[HelpNeeder]

u+6 = sin(t) + 8
<=>
du = cos(t)dt

But what do actually "du" and "dt" mean here?
Because we normally have d... in the numerator and d... in the denominator?

Thank you in advance.

 

[R.M.]

"du" and "dt" represent differentials, or very very very small changes in those variables. When you first learned slope, you may have heard the phrase "delta y over delta x", meaning change in one over the change in the other; in calculus, the "d" can be thought of as a delta that means change. Since calculus allows us to calculate things at an exact moment in time, the differential represents the idea of making an interval infinitesimally small (taking the limit as the size of the interval approaches 0). There are more rigorous and formal definitions/interpretations, so consider mine the "layman's terms" version.

 

[jalexandre]

Assuming you understand the concept of a derivative:

This is an implicitly defined equation, meaning you can not define one variable in terms of another. Have you learned about implicit differentiation?

On the left side of the equation, you can simply think of du as "derivative of u." On the right hand side, when you take the derivative of sin(t), you must use the chain rule. Sin(t) is a compound function, f(g(x)), where the "outside" function is the sine function and the "inside" function is simply t. When we take the derivative of a compound function, you take the derivative of the "outside" function and multiply it by the derivative of the "inside" function: f'(g(x))g'(x).

So, the derivative of sin(t) in this case is cos(t)dt, where cos(t) is the derivative of the "outside" and dt is the derivative of the "inside," which is t.

Long story short, du can be thought of as "derivative of u" and dt can be thought of as "derivative of t."

 

[HelpNeeder]

Thank you!
So what would du/dt in this case mean?

 

[Inertia_Squared]

The point is that those two values are generally arbitrary, but if you are trying to work with them, in this case, du would be your y-value and dt would be your x or time function (some trig functions will use time to represent the x-axis).

Keep in mind this is only a general statement and without the full question and working its hard to interpret what you are looking for. If you still haven't found out what you wanted please post the full question with working (if possible) and we can take a closer look!

 

[HelpNeeder]

Thank you!
The thing is that I work with derivatives a lot without fully understanding what these "d.."-s mean.
So du/dt = cos(t)
means that if the variable t changes with a very small value, then u changes with cos(t)?

 

[Inertia_Squared]

In short, yes. In long, there's more to it than that.

Dx can be treated as an expression, you could also say it is something like ([x+0.001]-x) and then then dy is the same except the y that you would get for putting those x's in the function. Let's put this into an example:

Say I have the function f(x) = 2x, then dy/dx could be written as (f(x+0.001)-f(x))/([x+0.001]-x)
The xs cancel and you get (f(x+0.001)-f(x))/(0.001)

So, with our function f(x) = 2x, let's make x = 3 and see what happens;

The gradient will equal [2(3.001)-2(3)]/0.001 which equals (6.002-6)/0.001 which is also 0.002/0.001 which is equal to 2, notice how the coefficient of x (the gradient) is the same as the gradient we just found? Now since a simple function like 2x already has its gradient known its not that helpful in this case, but when you get to functions with curves like x², it lets you find the gradient at any x value.

Im sure you already know all these things, but all the better, as the same applies for trig functions! If you're still having some trouble id recommend going through what I just did but with the function sinx, and you will notice if you keep plotting gradients on a graph that it actually makes a cosine graph, at your current level you shouldn't need to worry too much about why this is, just remember that it is!

 

[HelpNeeder]

Thank you very much for the explanation! I understand it much better now.

I was too busy applying formulas, so at one point I noticed that I didn't fully understand what I was doing. So I took a step back and thanks to your answers I refreshed the basics.

 

[HallsofIvy]

In short, yes. In long, there's more to it than that.

Dx can be treated as an expression, you could also say it is something like ([x+0.001]-x) and then then dy is the same except the y that you would get for putting those x's in the function. Let's put this into an example:

Say I have the function f(x) = 2x, then dy/dx could be written as (f(x+0.001)-f(x))/([x+0.001]-x)
The xs cancel and you get (f(x+0.001)-f(x))/(0.001)

No that's just wrong! That will give df/dx if f is a linear function but not any other functions. You are missing the concept of the "limit" which is essential to the derivative. \(\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}\). If h= 0.001 you might get an approximate value for the derivative but to get the actual derivative you have to use the limit.

So, with our function f(x) = 2x, let's make x = 3 and see what happens;

The gradient will equal [2(3.001)-2(3)]/0.001 which equals (6.002-6)/0.001 which is also 0.002/0.001 which is equal to 2, notice how the coefficient of x (the gradient) is the same as the gradient we just found? Now since a simple function like 2x already has its gradient known its not that helpful in this case, but when you get to functions with curves like x², it lets you find the gradient at any x value.

Im sure you already know all these things, but all the better, as the same applies for trig functions! If you're still having some trouble id recommend going through what I just did but with the function sinx, and you will notice if you keep plotting gradients on a graph that it actually makes a cosine graph, at your current level you shouldn't need to worry too much about why this is, just remember that it is!

but with a non-linear function, \(\displaystyle x^2\) for example, \(\displaystyle \frac{(x+ 0.001)^2- x^2}{0.001}= \frac{x^2+ 0.002x+ 0.00001- x^2}{0.001}= \frac{0.002x+ 0.00001}{0.001}= 2x+ 0.001\)
while the actual derivative is \(\displaystyle \lim_{h\to 0}\frac{(x+ h)^2- x^2}{h}= \lim_{h\to 0}\frac{x^2+ 2hx+ h^2- x^2}{h}= \lim_{h\to 0}\frac{2hx+ h^2}{h}= \lim_{h\to 0} 2x+ h= 2x\), NOT "2x+ 0.001"!

 

[Inertia_Squared]

No that's just wrong! That will give df/dx if f is a linear function but not any other functions. You are missing the concept of the "limit" which is essential to the derivative. \(\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}\). If h= 0.001 you might get an approximate value for the derivative but to get the actual derivative you have to use the limit.


but with a non-linear function, \(\displaystyle x^2\) for example, \(\displaystyle \frac{(x+ 0.001)^2- x^2}{0.001}= \frac{x^2+ 0.002x+ 0.00001- x^2}{0.001}= \frac{0.002x+ 0.00001}{0.001}= 2x+ 0.001\)
while the actual derivative is \(\displaystyle \lim_{h\to 0}\frac{(x+ h)^2- x^2}{h}= \lim_{h\to 0}\frac{x^2+ 2hx+ h^2- x^2}{h}= \lim_{h\to 0}\frac{2hx+ h^2}{h}= \lim_{h\to 0} 2x+ h= 2x\), NOT "2x+ 0.001"!

Yes I understand that, they said they were already familiar with differentiation however so I assumed it was implied to keep things simple. Thanks for the extra detail, though!