Master Calculus 1 Practice Problems with Expert Help

How to find instantaneous rate of change for $g(x)=x^2-<!--\; -\; -->x+4$ at x=-7?

The following distributional limit is well-known:
$\underset{a\to <!--\; \backslash rightarrow\; -->0^{+}}{lim}(\frac{1}{x-<!--\; -\; -->ia}-<!--\; -\; -->\frac{1}{x+ia})=2i\mathrm{\pi <!--\; \pi \; -->}\mathrm{\delta <!--\; \delta \; -->}(x).$

Smoothness properties of partial antiderivatives
I'm wondering what can be said about the well-behavior (or lack thereof) of antiderivatives of multivariable functions.
For example, consider the function $f(x,y)=2xy$. If we take the antiderivative of this function with respect to x, we obtain $F(x,y)=x^{2}y+h(y)$ where h(y) is an arbitrary function of y. The curious thing is there are no restrictions on what h could be. h could be a well-behaved differentiable function, or it could be a nasty function like the Weierstrass function, which is differentiable nowhere, or even the completely discontinuous rational number detector $h(x)=\{\begin{array}{ll}1& \text{if }x\text{ is rational}\\ 0& \text{if }x\text{ is irrational}\end{array}$
In either case, F would not be differentiable with respect to y, but even so, the partial derivative $\mathrm{\partial <!--\; \partial \; -->}F/\mathrm{\partial <!--\; \partial \; -->}x$ would be well-defined and will equal our original function f.
So obviously not every possible partial antiderivative will be differentiable with respect to the other variable, but clearly some are. If we choose $h(y)=y$, for example, then F is certainly differentiable with respect to y. In fact, as long as h is a differentiable function itself, F will be differentiable with respect to y.
My question is whether this holds true in general. If partial antiderivatives F exist for a multivariable function f with respect to one variable, is it always true that at least one such F will be differentiable with respect to the other variables?
I'm guessing the answer to this is No, but that raises the question of what requirements must we impose on the original function f to guarantee that a "nice" partial antiderivative exists. Is it enough that f be continuous? That f be fully differentiable? That f be continuously differentiable?
And even more generally, if I want to guarantee the existence of a partial antiderivative F that has some nice smoothness property (e.g. it is $C^2$, or $C^3$, etc.), what requirements must I impose on the original f to ensure this?

Evaluate the limit: ${\displaystyle \underset{(x,y)\to <!--\; \backslash rightarrow\; -->(0,0)}{lim}{\displaystyle \frac{x^2+y^2}{x^4+y^4}}}$

Calculate $\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}\frac{\sin <!--\; \; -->ax\cos <!--\; \; -->bx}{\sin <!--\; \; -->cx}$

How do you find the instantaneous rate of change of the function $f(x)=\ln <!--\; \; -->(2x)$ when x=3?

What does it mean to write an integral in its explicit form?
The first part of Fundamental Theorem of Calculus gives an abstract way of producing antiderivatives:
Write $F(x)={\int <!--\; \int \; -->}_1^x(t+1)dt$ in an explicit form. Check that its derivative is indeed $x+1$.
What does it mean to find an integral in it explicit form? Does it mean to find its anti derivative?

Evaluate $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$ (Making antiderivative continuous.)
Evaluate $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$
I know that this has an antiderivative on $\mathbb{R}$.
I can use the trig. substitution $t=\tan <!--\; \; -->x$ on $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->})$
$x=\arctan <!--\; \; -->t$
$dx=\frac{1}{1+t^2}dt$
To get $\int <!--\; \int \; -->\frac{1}{1+3{\sin }^2<!--\; \; -->x}dx$ = $\int <!--\; \int \; -->\frac{1}{1+3\frac{t^2}{t^2+1}}\cdot <!--\; \cdot \; -->\frac{1}{1+t^2}dt$= $\int <!--\; \int \; -->\frac{1}{1+(2t{)}^2}dt=\frac{1}{2}\int <!--\; \int \; -->\frac{1}{1+u^2}du=\frac{1}{2}\arctan <!--\; \; -->(2\tan <!--\; \; -->x)+C$
But this only appplies on the interval $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}+k\mathrm{\pi <!--\; \pi \; -->})$, whereas the original function should have an antiderivative on $\mathbb{R}$. So I have to make it continuous, but I don't know how.
(I do know that we did a similar thing for $\int <!--\; \int \; -->|x|dx$, but I lost my notes and don't know how doing this is called in english so I can't google it. We called it "gluing" antiderivatives. And I remember that we utilized limits in some fashion related to the integration constants)

Antiderivative 1/z on $\mathbb{C}$
Let $\mathrm{\Omega <!--\; \Omega \; -->}\subseteq <!--\; \subseteq \; -->\mathbb{C}$ be open and $\mathrm{\gamma <!--\; \gamma \; -->}:[\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}]\to <!--\; \to \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ be a piecewise continuously differentiable and closed path.
Why does $z^{-<!--\; -\; -->1}$ have no antiderivative on $\mathbb{C}\setminus <!--\; \setminus \; -->0$?
Why is $\underset{\mathrm{\gamma <!--\; \gamma \; -->}}{\int <!--\; \int \; -->}{z}^{-<!--\; -\; -->1}dz=2\mathrm{\pi <!--\; \pi \; -->}i\cdot <!--\; \cdot \; -->{\text{ind}}_{\mathrm{\gamma <!--\; \gamma \; -->}}(0)$, where ${\text{ind}}_{\mathrm{\gamma <!--\; \gamma \; -->}}$ denotes the winding number.

We have $f:\mathbb{R}\to <!--\; \backslash rightarrow\; -->\mathbb{R}$ and $a,b\in <!--\; \in \; -->\mathbb{R}$ such that:
$f(x)=a\cos <!--\; \; -->(|x^3-<!--\; -\; -->x|)+b|x|\sin <!--\; \; -->(|x^3+x|)$

Calculate limit involving the antiderivative of a function
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$, $f(x)=e^{x^2}$. Now, consider F an antiderivative of f.
Calculate: $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{xF(x)}{f(x)}$.
I couldn't find the antiderivative of f, and besides finding F, I have no other ideas.

How to following limit
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\frac{1}{n^2}+\frac{2}{n^2}+\cdots <!--\; \cdots \; -->+\frac{(n-<!--\; -\; -->1)}{n^2})(\frac{1}{n+1}+\frac{1}{n+2}+\cdots <!--\; \cdots \; -->+\frac{1}{n+n}).$

Show $(\genfrac{}{}{0ex}{}{n}{0})-<!--\; -\; -->\frac{1}{2}(\genfrac{}{}{0ex}{}{n}{1})+...+\frac{(-<!--\; -\; -->1{)}^n}{n+1}(\genfrac{}{}{0ex}{}{n}{n})=\frac{1}{n+1}$

Finding the Antiderivative of a complex function
I need to find the antiderivative of $f(z)=z\log <!--\; \; -->(z)$, but I am confused on how exactly to do that.
So we need to find $\int <!--\; \int \; -->z\log <!--\; \; -->(z)dz$ right? Since $z=x+iy$, then $\int <!--\; \int \; -->(x+iy)\log <!--\; \; -->(x+iy)(dx+idy)$, but then how do we split $\log <!--\; \; -->(x+iy)$?
Can we just integrate normally with respect to z THEN substitute with $z=x+iy$? We can use integration by parts as usual if so.

If
$f(x,y)=x^{x^{x^{x^y}}}+\ln <!--\; \; -->(x)[{\tan }^{-<!--\; -\; -->1}<!--\; \; -->({\tan }^{-<!--\; -\; -->1}<!--\; \; -->({\tan }^{-<!--\; -\; -->1}<!--\; \; -->(\sin <!--\; \; -->(\cos <!--\; \; -->xy)-<!--\; -\; -->\ln <!--\; \; -->(x+y)))]$

Confusion on ambiguity of antiderivatives.
Would this statement, $\mathrm{\forall <!--\; \forall \; -->}k\in <!--\; \in \; -->\mathbb{R}$, be true or false? And are these interchangeable under all circumstances?
$\int <!--\; \int \; -->f(x)dx+k=\int <!--\; \int \; -->f(x)dx$
Since we could assign F(x) to either the LHS or RHS, and differentiating F(x) would both result with f(x). However if we assign $I=\int <!--\; \int \; -->f(x)dx$, then $I=I+k\Rightarrow <!--\; \Rightarrow \; -->k=0$ even though we already stated "for all k". I think my misunderstanding stems from notation and what they represent. Would appreciate any clarification on this as I could not find a direct take on this flexibility of antiderivatives, especially in such context of equality.

I'm wondering about the motivation for proving the limit.
${\displaystyle \underset{x\to <!--\; \to \; -->1}{lim}\sqrt[n]{x}=1}$