Some questions while I self-learn Applied Calculus

[gmax137]

Here is some different letters

@user079622 are you aware that these derivatives you have copied here are "partial derivatives" (that's why they have the curly "d" as in: ${\partial f}$)? The normal nomenclature is:

$\frac {df} {dx}$, the derivative of function f with respect to x

$\frac {\partial f} {\partial x}$, the partial derivative of function f with respect to x

These are not the same, and if you do not know what a partial derivative is, that is OK, you just need to slow down and learn about "regular" derivatives first.

 

[user079622]

@user079622 are you aware that these derivatives you have copied here are "partial derivatives" (that's why they have the curly "d" as in: ${\partial f}$)? The normal nomenclature is:

$\frac {df} {dx}$, the derivative of function f with respect to x

$\frac {\partial f} {\partial x}$, the partial derivative of function f with respect to x

These are not the same, and if you do not know what a partial derivative is, that is OK, you just need to slow down and learn about "regular" derivatives first.

No, I haven't gotten to that part yet, I'm still searching what is the most effective way to learn...

 

[Mark44]

Learning from book is very hard, it is hard to figure out what you need to do with this "hieroglyphics".

What book are you using? As I mentioned before, some of what you call "hieroglyphics" is pretty much nonstandard, such as the notation $f_{'}$ to mean $\frac{\partial f}{\partial y}$ or $f_{''}$ to mean $\frac{\partial^2 f}{\partial y^2}$. As someone else and I already mentioned, these are all partial derivatives. It looks to me like you have skipped too far ahead in this book.

just one example:
In book I cant find how they get du, because book skips steps.

What's the example from the book? The first example I saw in the Youtube video seemed pretty obvious.

 

[user079622]

What book are you using? As I mentioned before, some of what you call "hieroglyphics" is pretty much nonstandard, such as the notation $f_{'}$ to mean $\frac{\partial f}{\partial y}$ or $f_{''}$ to mean $\frac{\partial^2 f}{\partial y^2}$. As someone else and I already mentioned, these are all partial derivatives. It looks to me like you have skipped too far ahead in this book.

What's the example from the book? The first example I saw in the Youtube video seemed pretty obvious.

Book for math I for engineering university on my mother language.

I just read text about u-substitution, in every book they dont show all steps..So for me it easier to watch video where professor solve tasks with u-substitution.

 

[user079622]

@Mark44

Find integral of sin 2x dx
step 1. substitution
2x=t
x=t/2 differentiate........dx= dt/ 2

Why x become dx and t become dt, if derivation of x is 1, (x)´=1 ?

 

[Mark44]

Find integral of sin 2x dx
step 1. substitution
2x=t
x=t/2 differentiate........dx= dt/ 2

The letter u is often used for substitutions like this.
Let u = 2x.
Then du = 2 dx.
So $u = 2x \Rightarrow x = \frac u 2$, and $dx = \frac {du} 2$.

Then the integral $\int \sin(2x)dx$ becomes $\int \sin(u) \frac{du}2 = \frac 1 2 \int \sin(u) du$.
Can you carry out the remainder of this problem?

To answer your question...

Why x become dx and t become dt, if derivation of x is 1, (x)´=1 ?

With substitutions you need to work with differentials, not derivatives.
The differential of x is dx. If u is a function of x, then $du = \frac {du}{dx}\cdot dx$.
Here u = 2x, so $du = 2 \cdot dx$.

 

[user079622]

Then the integral $\int \sin(2x)dx$ becomes $\int \sin(u) \frac{du}2 = \frac 1 2 \int \sin(u) du$.
Can you carry out the remainder of this problem?

Yes
=-cos (u) x 1/2 + C= -cos (2x) /2 +C

With substitutions you need to work with differentials, not derivatives.
The differential of x is dx. If u is a function of x, then $du = \frac {du}{dx}\cdot dx$.
Here u = 2x, so $du = 2 \cdot dx$.

only after watching this video do I understand what it is differential

 

[PeroK]

With substitutions you need to work with differentials, not derivatives.

Which is, of course, true of the technique normally used for substitutions. Just this morning, however, I showed how this technique is related to the integration by change of variables, which does not depend on differentials!

https://www.physicsforums.com/threads/question-about-change-of-variables.1060606/#post-7065893

 

[Mark44]

only after watching this video do I understand what it is differential

In every calculus textbook I've ever seen (and there are many), the technique of integration by substitution is almost the first method shown for evaluating integrals. All of the examples show what the substitution is, and from it how to replace the integrand and dx. Somewhere after derivatives are presented, there is usually some discussion of differentials. You didn't mention the title of your calculus textbook or its authors, but I'm starting to think that it isn't a very good source for learning calculus.

 

[Mark44]

I showed how this technique is related to the integration by change of variables, which does not depend on differentials!

From that post, you wrote

The change of variables is really the inverse of the chain rule. Formally, if $f$ and $u$ are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$

Although differentials aren't explicitly mentioned, the substitution evidently is t = u(x), so dt on the right side corresponds to u'(x) dx on the left side of the equation above,

 

[PeroK]

Although differentials aren't explicitly mentioned, the substitution evidently is t = u(x), so dt on the right side corresponds to u'(x) dx on the left side of the equation above,

That observation motivates the idea of differentials, in terms of something that might makes sense outside the integrand. And, motivates the substitution technique. The proof of the change of variables formula, however, involves only the chain rule and the FTC. It doesn't involve an appeal to dfferentials or infinitesimals. Which is an important point, IMO.

 

[user079622]

I will report back in a few days when I pass the differentials.