#### HelpNeeder

##### New member

- Joined
- Feb 17, 2021

- Messages
- 5

<=>

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.

- Thread starter HelpNeeder
- Start date

- Joined
- Feb 17, 2021

- Messages
- 5

<=>

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.

- Joined
- Feb 4, 2021

- Messages
- 5

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."

- Joined
- Feb 17, 2021

- Messages
- 5

Thank you!

So what would du/dt in this case mean?

So what would du/dt in this case mean?

- Joined
- May 24, 2019

- Messages
- 51

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!

- Joined
- Feb 17, 2021

- Messages
- 5

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)?

- Joined
- May 24, 2019

- Messages
- 51

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 x

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!

- Joined
- Feb 17, 2021

- Messages
- 5

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.

- Joined
- Jan 27, 2012

- Messages
- 7,164

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.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)

but with a non-linear function, \(\displaystyle x^2\) for example, \(\displaystyle \frac{(x+ 0.001)^2- x^2}{0.001}= \frac{x^2+ 0.002x+ 0.00001- x^2}{0.001}= \frac{0.002x+ 0.00001}{0.001}= 2x+ 0.001\)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 x^{2}, 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!

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"!

- Joined
- May 24, 2019

- Messages
- 51

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!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"!