Calculus 2 Equations: Expert Solutions and More

Prove that if $f$ is defined for $|x|<r$ and if there exists a constant $B$ such that
$|f^n(x)|\le <!--\; \le \; -->B$
for all $|x|<r$ and $n\in <!--\; \in \; -->\mathbb{N}$, then the Taylor series expansion :
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{f^{(n)}(0)x^n}{n!}$
converges to $f(x)$ for $|x|<r$.

Let $f$ be continuous on $[0,1]$ and $f(0)=f(1)$. Also, $n\in <!--\; \in \; -->\mathbb{N}$. Prove that there exists some $x$ s.t. $f(x)=f(x+\frac{1}{n})$.
I think I need to assume towards a contradiction that if $g(x)=f(x)-<!--\; -\; -->f(x+\frac{1}{n})$, then $g(x)\ne <!--\; \ne \; -->0$ $\mathrm{\forall <!--\; \forall \; -->}x$.
So, $g(0)=f(0)-<!--\; -\; -->f(\frac{1}{n})$, $g(1)=f(1)-<!--\; -\; -->f(1+\frac{1}{n})$. Since $f(0)=f(1)$,
$g(0)-<!--\; -\; -->g(1)=f(1+\frac{1}{n})-<!--\; -\; -->f(\frac{1}{n})$. For context, I am only up to Ch. 8 in Spivak, so I cannot use any derivative-based theorems which is why I am stuck here.

What is the derivative of ${\displaystyle x=\tan (x+y)}$

The sum of the square of a positive number and the suare of 3 more than the number is 29. What is the number?

When I check my answer using the some website, I get a result I can't agree with. So I'd like to hear your opinion :)
Asked: use implicit differentiation to differentiate $\sin <!--\; \; -->(xy)$.
My take on the matter:
Using the chain rule: $\frac{d}{dx}\sin <!--\; \; -->(xy)$ = $\cos <!--\; \; -->(xy)\frac{d}{dx}xy$, then using the product rule on latter factor we get: $\frac{d}{dx}xy=y+x\frac{dy}{dx}$
Hence: $\frac{d}{dx}\sin <!--\; \; -->(xy)=y\cos <!--\; \; -->(xy)+x\cos <!--\; \; -->(xy)\frac{dy}{dx}$.
Is this correct?

I'm trying to solve the below exercise.

Let $f:[0,1]\to <!--\; \to \; -->[0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a\in <!--\; \in \; -->A$ such that $f(a)=a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0)=0$ or $f(1)=1$, the theorem is proved. Suppose not. Define $g(x)=f(x)-<!--\; -\; -->x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0)\ne <!--\; \ne \; -->0$, $g(0)\ne <!--\; \ne \; -->0$, so $g(0)>0$. Since $f(1)\ne <!--\; \ne \; -->1$, $g(1)<0$. By the intermediate value theorem, there exists $a\in <!--\; \in \; -->[0,1]$ such that $g(a)=0$. So $f(a)-<!--\; -\; -->a=0$, so $f(a)=a$.
I am fairly sure that the result is not true for discontinuous functions, largely because I requiblack continuity of $g$ to invoke the intermediate value theorem. I am having trouble finding a counterexample, however. Do I define a function piece-wise, with jumps at $x=0$ or $x=1$, to try to break the intermediate value theorem?

In higher dimensions, is the derivative (jacobians,gradients etc.) defined using taylor series or taylor series formula proved through derivatives ?

How do you differentiate ${\displaystyle x{y}^2-x^{3}y=6}$

Prove that :
$\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^k}{(2k+1{)}^3}=\frac{{\mathrm{\pi <!--\; \pi \; -->}}^3}{32}.$

4xy^3+5xy=27 Point:(3,1)
So, using Implicit Differentiation...
(4y^3+4xy^2(dy/dx))+(5y+5x(dy/dx))=0
4xy^2(dy/dx)+5x(dy/dx)=-4y^3-5y
dy/dx(4xy^2-5x)=-4y^3-5y
dy/dx=(-4y^3-5y)/(4xy^2+5x)
Now I replace x and y in the equation with x=3 and y=1
I get 20/27, which is apparently incorrect. What am I doing wrong?

I'm supposed to implicitly differentiate the following and give the answer in terms of $y\mathrm{\prime <!--\; \prime \; -->}$
$\tan <!--\; \; -->(x-<!--\; -\; -->y)=\frac{y}{1+x^2}$
$\frac{(1+x^2)y\mathrm{\prime <!--\; \prime \; -->}-<!--\; -\; -->2xy}{(1+x^2{)}^2}$
How do I solve for $y\mathrm{\prime <!--\; \prime \; -->}$?
Edit: After a ludicrous amount of algebra, i finally ended up at
$\frac{se{c}^2(x-<!--\; -\; -->y)(1+x^2{)}^2+2xy}{(1+x^2)(1+se{c}^2(x-<!--\; -\; -->y)(1+x^2))}$
which apparently is correct.

Calculating Rate of Change
At the point (0,1,2) in which direction does the function $f(x,y,z)=x{y}^{2}z$ increase most rapidly? What is the rate of change of $f$ in this direction? At the point (1,1,0), what is the derivative of $f$ in the direction of the vector $2\stackrel{^<!--\; ^\; -->}{i}+3\stackrel{^<!--\; ^\; -->}{j}+6\stackrel{^<!--\; ^\; -->}{k}$?
I assumed that the rate of change is the same as the gradient of the function, namely $▽<!--\; ▽\; -->f$. Calculating this gave me:
$▽<!--\; ▽\; -->f=\frac{\mathrm{\partial <!--\; \partial \; -->}(x{y}^{2}z)}{\mathrm{\partial <!--\; \partial \; -->}x}\stackrel{^<!--\; ^\; -->}{i}+\frac{\mathrm{\partial <!--\; \partial \; -->}(x{y}^{2}z)}{\mathrm{\partial <!--\; \partial \; -->}y}\stackrel{^<!--\; ^\; -->}{j}+\frac{\mathrm{\partial <!--\; \partial \; -->}(x{y}^{2}z)}{\mathrm{\partial <!--\; \partial \; -->}z}\stackrel{^<!--\; ^\; -->}{k}$
$=y^{2}z\stackrel{^<!--\; ^\; -->}{i}+2xz\stackrel{^<!--\; ^\; -->}{j}+x{y}^2\stackrel{^<!--\; ^\; -->}{k}$
Evaluating at point:
$▽<!--\; ▽\; -->f(0,1,2)=2\stackrel{^<!--\; ^\; -->}{i}$
Hence, the function increases most rapidly in the x direction.
I am uncertain of how to approach solving the third part of the question, should I evaluate the rate of change at (1,1,0) and then find the difference between that and the vector $2\stackrel{^<!--\; ^\; -->}{i}+3\stackrel{^<!--\; ^\; -->}{j}+6\stackrel{^<!--\; ^\; -->}{k}$?

I found that $f(x)=x^2$ is continuous on $[1,2]$ but does not assume the values 2 intermediate between the value 1 and 4. I do not understand what they mean. For me, $\sqrt{2}\in <!--\; \in \; -->[1,2]$ and $f(\sqrt{2})=2$.
Also, using the the intermediate value theorem, since $f$ is continuous on $[1,2]$ and $f(1)=1<2<4=f(2)$, there exists $c\in <!--\; \in \; -->[1,2]$ such that $f(c)=2$.
Can I get some explanation about this?

Are there instances when a Taylor series and a Laurent series of the same function about the same point ever equal?

Integrate the Taylor series
$e^{(-<!--\; -\; -->t^2)}=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->t^2{)}^n}{n!}$
term-by-term to obtain the Taylor series for erf (error function) about $a=0$.

find $\frac{dy}{dx}$ of $x^3+y^3+8$ using partial derivative
So what I need to do is $\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}$ and $\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}$ which are:
$3x^2$ and $3y^2$ and than to divide to get $\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}}{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}}=\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=1$?

Test the convergences of the following series
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^n\frac{\sin <!--\; \; -->n}{\sqrt{n}}$

Let $F(x)$ be the unique function that satisfies $F(0)=0$ and $F\prime (x)={\displaystyle \frac{\sin <!--\; \; -->(x^3)}{x}}$ for all $x$. Find the Taylor series for $F(x)$ about $x=0$.
Wouldn't the value of every derivative at $0$ just be $0$ ? So how does a Taylor series even exist?
If $F(0)=0$, then can the Taylor series of $F(x)$ be the same as that of $F^{\prime }(x)$ ?

Use a table of integrals to evaluate the following integrals. $$\int <!--\; \int \; -->{\cos }^3<!--\; \; -->xdx$$

How would you differentiate
$\sec <!--\; \; -->x{e}^x$