Master Calculus 1 Practice Problems with Expert Help

Compute the derivative of the function F(t)
$F(t)=\iint <!--\; \iint \; -->e^{\frac{tx}{y}}dxdy$
where $0.1\le <!--\; \le \; -->x\le <!--\; \le \; -->t$
and $0.1\le <!--\; \le \; -->y\le <!--\; \le \; -->t$
Please help me to find F'(t)
I tried the antiderivatives. But F(t) doesn't seem to be an elementary integral.

Derivative of the integral of composite functions?
$L=\int <!--\; \int \; -->f(g(x))dx$, what is ${\displaystyle \frac{dL}{dg(x)}}$?

On what integration domain is the author finding the anti derivative?
Find the antiderivative(s): $\int <!--\; \int \; -->\frac{1}{x^2-<!--\; -\; -->a^2}\mathbb{d}x$.
$\int <!--\; \int \; -->\frac{1}{x^2-<!--\; -\; -->a^2}\mathbb{d}x$
$\int <!--\; \int \; -->\frac{1}{(x+a)(x-<!--\; -\; -->a)}\mathbb{d}x$
$\frac{1}{2a}\int <!--\; \int \; -->\left(\frac{1}{x-<!--\; -\; -->a}-<!--\; -\; -->\frac{1}{x+a}\right)\mathbb{d}x$
$\frac{1}{2a}[\ln <!--\; \; -->(x-<!--\; -\; -->a)-<!--\; -\; -->\ln <!--\; \; -->(x+a)]+C$
$\frac{1}{2a}\ln <!--\; \; -->(|\frac{x-<!--\; -\; -->a}{x+a}|)+C,x\ne <!--\; \ne \; -->\pm <!--\; \pm \; -->a$
$\frac{1}{x^2-<!--\; -\; -->a^2}$ is a discontinuous function, and there are three integration domains of it: $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->a)$, $(-<!--\; -\; -->a,a)$ and $(a,+\mathrm{\infty <!--\; \infty \; -->})$. We will get three different family of functions as antiderivatives for the three different integration domains of this function, and two antiderivatives belonging to two different integration domains/family of functions will not differ by a constant.
My question is that my book found out the antiderivative belonging to which integration domain?

A sharp Stirling's approximation form states that
$n!\sim <!--\; \sim \; -->{\left(\frac{n}{e}\right)}^n\sqrt{2\mathrm{\pi <!--\; \pi \; -->}n}.$
Use that form to show that:
$(\genfrac{}{}{0ex}{}{2m}{m})=\mathrm{\Theta <!--\; \Theta \; -->}\left(\frac{2^{2m}}{\sqrt{m}}\right).$

What is an antiderivative and can 1 function have more than one antiderivatives?
I want to know what the definition of an antiderivative is really. I know that the antiderivative of a function between 2 limits will give you the area under the graph between these limits. Take the function $y=1/(-<!--\; -\; -->2x)$. The integral will be $-<!--\; -\; -->(1/2)\ln <!--\; \; -->|x|$ or it can also be $-<!--\; -\; -->(1/2)\ln <!--\; \; -->(2|x|)$ which is equal to $-<!--\; -\; -->0.5\ln <!--\; \; -->(2)-<!--\; -\; -->0.5\ln <!--\; \; -->|x|$. So as you can see they differ only by a constant. So when you plug in two limits you will get the same value. But i am wondering what an antiderivative represents exactly and how can it be possible as you saw to have a function with 2 antiderivatives. And by the way, can an antiderivative be represented by a graphical method?

Antiderivative of $\frac{1-<!--\; -\; -->x^n}{1-<!--\; -\; -->x}$
Is there a simple antiderivative of $\frac{1-<!--\; -\; -->x^n}{1-<!--\; -\; -->x},n\in <!--\; \in \; -->\mathbb{N}$?
This functions seems realy easy, but if I try the product rule, I get a worse fraction.

How can I find the constants A and B if
$\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\sqrt{x^2+x}-<!--\; -\; -->Ax-<!--\; -\; -->B)=0?$

Limit of difference quotient of $\frac{t^4}{2}+t^3-<!--\; -\; -->\frac{t^2}{2}+1$

$\underset{i=1}{\overset{k-<!--\; -\; -->1}{\prod <!--\; \prod \; -->}}(1-<!--\; -\; -->\frac{i}{2N})\approx 1-<!--\; -\; -->\frac{(\genfrac{}{}{0ex}{}{k}{2})}{2N}$
what assumption are needed for this approximation to be useful.

Use the linear approximation to $f(x,y)=\sqrt{\frac{x+5}{y+2}}$ at $(2,3)$ to estimate $\sqrt{\frac{7.1}{4.9}}$.

I know that the above limit is 1 from Wolfram but I had no proper way to prove. I attempted a proof and it is as follows:
${\displaystyle \underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left({\displaystyle \frac{1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+\cdots <!--\; \cdots \; -->+{\displaystyle \frac{1}{n}}}{\log <!--\; \; -->n}}\right)={\displaystyle \frac{{\displaystyle {\int <!--\; \int \; -->}_0^1{\displaystyle \frac{1}{n}}dn}}{{\displaystyle \underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\log <!--\; \; -->n}}}.}$

Double angle formulae for $\sin <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}$ and $\cos <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}$ in tangent form?

I want to determine the limit of the following sequence
$x_n=\frac{\frac{2}{1}+\frac{3^2}{2}+\frac{4^3}{3^2}+...+\frac{(n+1{)}^n}{n^{n-<!--\; -\; -->1}}}{n^2}$

Textbooks that use notation with explicit argument variable in the upper bound ${\int <!--\; \int \; -->}^x$ for "indefinite integrals."
I dare to ask a question similar to a closed one but more precise.
Are there any established textbooks or other serious published work that use ${\int <!--\; \int \; -->}^x$ notation instead of $\int <!--\; \int \; -->$ for the so-called "indefinite integrals"?
(I believe I've seen it already somewhere, probably in the Internet, but I cannot find it now.)
So, I am looking for texts where the indefinite integral of cos would be written something like:
${\int <!--\; \int \; -->}^x\cos <!--\; \; -->(t)dt=\sin <!--\; \; -->(x)-<!--\; -\; -->C$
or ${\int <!--\; \int \; -->}^x\cos <!--\; \; -->(x)dx=\sin <!--\; \; -->(x)+C.$
(This notation looks more sensible and consistent with the one for definite integrals than the common one with bare $\int <!--\; \int \; -->$.)
IMO, the indefinite integral of f on a given interval I of definition of f should not be defined as the set of antiderivatives of f on I but as the set of all functions F of the form
$F(x)={\int <!--\; \int \; -->}_a^{x}f(t)dt+C,x\in <!--\; \in \; -->I,$,
with $a\in <!--\; \in \; -->I$ and C a constant (or as a certain indefinite particular function of such form). In other words, I think that indefinite integrals should be defined in terms of definite integrals and not in terms of antiderivatives. (After all, the integral sign historically stood for a sum.)
In this case, the fact that the indefinite integral of a continuous function f on an interval I coincides with the set of antiderivatives of f on I is the contents of the first and the second fundamental theorems of calculus:
1. The first fundamental theorem of calculus says that every representative of the indefinite integral of f on I is an antiderivative of f on I, and
2. The second fundamental theorem of calculus says that every antiderivative of f on I is a representative of the indefinite integral of f on I (it is an easy corollary of the first one together with the mean value theorem).

I'm trying to use L'Hopital's rule to calculate:
$\underset{x\to <!--\; \backslash rightarrow\; -->0^{+}}{lim}{\displaystyle \frac{x-<!--\; -\; -->\sin <!--\; \; -->x}{(x\sin <!--\; \; -->x{)}^{(3/2)}}}$

Antiderivatives and definite integrals - 2
Prove that the function $f(x)=x^2{\int <!--\; \int \; -->}_0^{1}t{\sin }^2<!--\; \; -->(tx)dt$
is differentiable in $\mathbb{R}$ and determine the formula f′(x) of its derivative.

Does the limit exist
$\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{n-<!--\; -\; -->H_n-<!--\; -\; -->\underset{j\le <!--\; \le \; -->n}{\sum <!--\; \sum \; -->}\frac{\mathrm{\Omega <!--\; \Omega \; -->}\left(j\right)}{j}+\underset{j\le <!--\; \le \; -->n}{\sum <!--\; \sum \; -->}\frac{\mathrm{\omega <!--\; \omega \; -->}\left(j\right)}{j}}{n}?$

Derivatives of Sets
So, I often see this:
$D_x[\int <!--\; \int \; -->f(x)]=f(x)$
But this is a derivative of a set of antiderivatives. What is the conceptual backing for this i.e. what does it mean to derive a set?