Calculus 2 Equations: Expert Solutions and More

For each $x\in <!--\; \in \; -->[0,10)$, write its decimal expansion $x=a_0.a_1{a}_2{a}_3\dots <!--\; \dots \; -->a_n\dots <!--\; \dots \; -->$. Take into account the number $x^{\prime }$ formed by the digits of odd rank of this decimal expansion; i. e., $x^{\prime }=0.a_1{a}_3{a}_5\dots <!--\; \dots \; -->a_{2n-<!--\; -\; -->1}\dots <!--\; \dots \; -->$ We have two cases:
(1) $x^{\prime }$ is not periodic
(2) $x^{\prime }$ is periodic after the digit $a_{(2n_0-<!--\; -\; -->1)}$
Now let $f$ be the function from the half-open interval [0,10) to $\mathbb{R}$ defined by
$f(x)=\{\begin{array}{ll}0& \text{in case (1) for }{x}^{\prime }\\ a_{(2n_0)}.a_{2(n_0+1)}\dots <!--\; \dots \; -->a_{2(n_0+k)}\dots <!--\; \dots \; -->& \text{in case (2) for }{x}^{\prime }\end{array}$
(Note that in case (2), the integer part $[f(x)]=a_{(2n_0)}$ of $f(x)$ can be nonzero.)

QUESTION
Prove that for all $x_1,x_2$, the function $f$ takes on the open interval $(x_1,x_2)$ all values between $f(x_1)$ and $f(x_2)$ despite $f$ being discontinuous in all points of its domain.

I want to solve $dy/dx$ for the following:
$x^2+y^2=R^2$ where $R$ is a constant.
I know to use implicit differentiation, though I have a question. When I derive $R^2$, do I obtain $2R$ or 0?
Additionally, deriving $y^2$ with respect to x yields $2y(dy/dx)$? This is different from a partial derivative?
Thanks!

How do you differentiate ${\displaystyle \tan 2x=\cos 3y}$

My textbook Elementary Classical Analysis claims that by Darboux's theorem (the intermediate value theorem for derivatives), if a function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ has a nonzero derivative on $\mathbb{R}$, then is $f$ strictly monotonic (i.e., either $f^{\prime }(x)>0$ on $\mathbb{R}$ or $f^{\prime }(x)<0$ on $\mathbb{R}$).
The claim is definitely true if $f^{\prime }s$ domain were instead a closed interval, but since $\mathbb{R}$ is open, I don't understand why Marsden's claim should be true.

The Taylor series as: $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{f^{(n)}(a)}{n!}(x-<!--\; -\; -->a{)}^n$
where $a=0$ is the Maclaurin series.
The Maclaurin series for $\frac{1}{1-<!--\; -\; -->x}$ is the geometric series $1+x+x^2+...$
what is $f(a)$ for the Taylor Series that results in this MacLaurin series?
then goes into the Taylor series for $\frac{1}{x}$ at $a=1$ is $1-<!--\; -\; -->(x-<!--\; -\; -->1)+(x-<!--\; -\; -->1{)}^2-<!--\; -\; -->(x-<!--\; -\; -->1{)}^3+...$
how this is derived?

How do you implicitly differentiate ${\displaystyle x^2+2xy-y^2+x=2}$

How do you find the derivative of ${\displaystyle x^{2}y}$

So I'm trying to differentiate an equation using implicit differentiation.
I start with $e^{x/y}=7x-<!--\; -\; -->y$
Now the left side of the eqn is where I'm having trouble.
I tried to use differentiation rules for exponentials, but this is incorrect.
Here's what I tried though:
$(e^x{)}^{1/y}\ln <!--\; \; -->(e^x)y^{\prime }=7-<!--\; -\; -->y^{\prime }$
simplfied to:
$x(e^x{)}^{1/y}\cdot <!--\; \cdot \; -->y^{\prime }=RS$

How to get the proper maclaurin series representation for $\cos <!--\; \; -->x$

So my problem is that I have to use implicit differentiation to find the derivative of $F(x,y)=e^{xy}-<!--\; -\; -->x$ when $F(x,y)=10$ and the equation of the tangent at the point $(1,\log <!--\; \; -->(11))$. So I tried to solve this using two ways:
The first way I used was rearranging the equation to $10+x=e^{xy}$ and then using $\ln$ to simply into $\ln <!--\; \; -->(10+x)=xy$ before using implicit differentiation. The result is
$\frac{dy}{dx}=\frac{1}{10x+x^2}-<!--\; -\; -->\frac{y}{x}$
and after substituting the point, I get approximately −0.950 as the slope.
The second way I used was just differentiating it straight away rather than rearranging and I get
$\frac{dy}{dx}=\frac{\frac{1}{e^{xy}}-<!--\; -\; -->y}{x}.$
But when I sub the point in I get approximately −0.688 as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?

find Maclaurin series for function $f(x)=\frac{1}{x^2+x+1}$

Considering s is implicit to function of $p$, given by $s^6-<!--\; -\; -->p^4=1$. For what s is it increasing and decreasing?
Well, I answeblack first like following: Calculating the first derivative using the implicit differentiation is $s^{\prime }=\frac{2p^3}{3s^5}$. $s$ is increasing where $s^{\prime }>0$ and decreasing where $s^{\prime }<0$.
Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

There is no trouble when I have to differentiate a function $f(x,y)$ where $y=g(x)$ using Implicit Function Theorem. It's easy to get $g_{xx}$ and $g_{yy}$. For instance, the formula for $g_{xx}=-<!--\; -\; -->\frac{f_{xx}{F}_y^2-<!--\; -\; -->(F_{xy}+F_{yx})F_x{F}_y+F_{yy}{F}_x^2}{F_y^3}$ knowing that $g^{\prime }(x)=-<!--\; -\; -->\frac{F_x}{F_y}$ whenever $F_y\ne <!--\; \ne \; -->0$.
Now, i'm in the situation $z+x+(y+z{)}^4=0$ where it defines an implicit function $z=f(x,y)$.
I got $f_x$ and $f_y$ using the theorem. However, I can't get $f_{xx}$ and $f_{yy}$ from there. I know I shoud apply the chain rule but I get lost whenever trying to get a "general formula" for them.

The question asks to implicitly differentiate $\tan <!--\; \; -->(x+y)=x$.
I get the correct answer of $-<!--\; -\; -->{\sin }^2<!--\; \; -->(x+y)$. However, the book also says "or you can use $-<!--\; -\; -->{\displaystyle \frac{x^2}{x^2+1}}$". I don't see how I can construct a proper right triangle to give/obtain this equivalent answer. Can you assist?

Suppose that I have a polynomial of degree 4 like $x^4+4x+c=0$ where $c\in <!--\; \in \; -->\mathbb{R}$.
What is the difference when the question says
Show that it has at most two real roots
and
Show that it has exactly two real roots
?
Are approaches to solve the problem in these two cases different?

Rate of change vs. average rate of change of a function
What is the difference between the both? I am trying to find a separate definitions for both in a textbook, but it only defines average rate of change.
I know that average rate of change is:
$\frac{changeiny}{changeinx}$
But what is just rate of change then?

When should I use total differentials and when should I use implicit differentiation?
An example of an equation where I'm unsure if to use total differentials or implicit differentiation is $f(x,y)=x^{2}y$
Using total differentials
$\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}dx+\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}dy=2xydx+x^{2}dy$, solving for $\frac{dy}{dx}=\frac{-<!--\; -\; -->2y}{x}$
Using implicit differentiation:
$(2x\frac{dy}{dx})(y)+x^2$, solving for $\frac{dy}{dx}=\frac{-<!--\; -\; -->x}{y}$
I'm unsure when to use each method and they both yield different results.

I need help finding the implicit differentiation of $x^4+\sin <!--\; \; -->y=x^3{y}^2$ I especially need help finding ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}$ for both sides. I'm having trouble with the product rule on the right side with ${\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$ of $x^3{y}^2$.

$\frac{1}{2\cdot <!--\; \cdot \; -->4}+\frac{1\cdot <!--\; \cdot \; -->3}{2\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->6}+\frac{1\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->5}{2\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->6\cdot <!--\; \cdot \; -->8}+\frac{1\cdot <!--\; \cdot \; -->3\cdot <!--\; \cdot \; -->5\cdot <!--\; \cdot \; -->7}{2\cdot <!--\; \cdot \; -->4\cdot <!--\; \cdot \; -->6\cdot <!--\; \cdot \; -->8\cdot <!--\; \cdot \; -->10}+\cdots <!--\; \cdots \; -->$
is equal to?