Calculus 2 Equations: Expert Solutions and More

Find the Taylor polynomial of degree n=4 for each function expanded about the given value of $x_0$.
$f(x)=\sqrt{x},x_0=1$

Find $\frac{dy}{dx}$ for $y=x^{-<!--\; -\; -->2/9}$

Solve the given differential equation subject to the indicated initial condition.
$(1+x^4)dy+x(1+4y^2)dx=0,y(1)=0$

Solve the following initial value problem:
$(1-<!--\; -\; -->t^2)y^t-<!--\; -\; -->4ty=1,y(2)=-<!--\; -\; -->1$
What is the largest interval on which its solution is guaranteed to uniquely exist?

I was looking to implicitly differentiate
$-<!--\; -\; -->22x^6+4x^{33}y+y^7=-<!--\; -\; -->17$
and found it to be
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{132x^5-<!--\; -\; -->132x^{32}y}{4x^{33}+7y^6}}$
Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with
${\displaystyle \frac{0}{11}}$
Now I go to solve
$y-<!--\; -\; -->y1=m(x-<!--\; -\; -->x1)$
getting
$y-<!--\; -\; -->1=0(x-<!--\; -\; -->1)$
resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?

How do you use implicit differentiation to find ${\displaystyle \frac{d^{2}y}{{dx}^2}}$ of ${\displaystyle x^3+y^3=1}$

I'm stuck on this problem involving implicit differentiation.
The instructions ask me to find $y^{\prime }$, the problem is:
$(x-<!--\; -\; -->2y{)}^3=2y^2-<!--\; -\; -->3$
So far I've been able to get this far:
$3(x-<!--\; -\; -->2y{)}^2(1-<!--\; -\; -->2y^{\prime })=4y(y^{\prime })$
I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.

(d) Let $f:[a,b]\to <!--\; \to \; -->[a,b]$ be a continuous function. Prove that $f$ has a fixed point, i.e. that there is a $с\in <!--\; \in \; -->[a,b]$ such that $f(c)=c$.
In the solution, it says that $f(a)\ge <!--\; \ge \; -->a$ and $f(b)\le <!--\; \le \; -->b$ but it do not seem obvious for me. If I am just given that a $f:[a,b]\to <!--\; \to \; -->[a,b]$, how do I know is this function increasing, decreasing or just a horizontal line?

Find the Maclaurin series for $f(x)=(x^2+4)e^{2x}$ and use it to calculate the 1000th derivative of $f(x)$ at $x=0$. Is it possible to just find the Maclaurin series for e2x and then multiply it by $(x^2+4)$?

I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:
Suppose $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->\mathbb{R}$ is continuous and satisfies $f(-<!--\; -\; -->1)=f(1)$. Prove that there exists $y\in <!--\; \in \; -->[0,1]$ such that $f(y)=f(y-<!--\; -\; -->1)$.
So far, I have consideblack a new function $g(x)=f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$, $x\in <!--\; \in \; -->[0,1]$, but am stuck after this.

The temperature $T(x)$ at each point $x$ on the surface of Mars (a sphere) is a continuous function. Show that there is a point $x$ on the surface such that $T(x)=T(-<!--\; -\; -->x)$
(Hint: Represent the surface of Mars as $\{x\in <!--\; \in \; -->{\mathbb{R}}^3:||x||=1\}$.)
Consider the function $f(x)=T(x)-<!--\; -\; -->T(-<!--\; -\; -->x)$
So.....
I consider an unit sphere is locating at the origin of a $xyz$-plane.
As $||x||=1$, I can say with $radius=1=\sqrt{x^2+y^2+z^2}$
To find there is a point $T(x)=T(-<!--\; -\; -->x)$, we use the formula $f(x)=T(x)-<!--\; -\; -->T(-<!--\; -\; -->x)$ and show somehow $f(x)$ will equal to 0 ???
It will be a point in the upper hemisphere and another point with the exactly opposite vector (if using $ijk$ plane) on the lower hemisphere

Why, after differentiating $y$ on one side of the equation, is $dy/dx$ added?
As clarification an example I will provide an example:
Implicitly differentiate $y^2=x$.
You get
$2y\frac{dy}{dx}=1$
as one of the first steps in differentiation. Why is the $dy/dx$ added after $y^2$ is differentiated?

Suppose that $x$ is a solution to the initial value problem $x^{\prime }=\frac{x^3-<!--\; -\; -->x}{1+t^2{x}^2}$ and $x(0)=\frac{1}{2}$ Show that $0<x(t)<1$ for all $t$ for which $x$ is defined.
For this question, I am told to use proof by contradiction. First find constant solutions, and then use intermediate value theorem to argue that since constant solution and initial value solution don't cross, by uniqueness theorem their existence contradict each other, hence $0<x(t)<1$. Can anyone explain how intermediate value theorem plays an effect in here?

Find,
$x^2{y}^2-<!--\; -\; -->2xy+x=2$
find the slope and the equation of the tangent line at (2,0) by implicit differentiation.
Part of the problem is that my text book says find $d/dx$ and then ignore $dy/dx$ after I get
$x^2{y}^2=2x^{2}y$
Please show me where I am going wrong. I would be able to find the equation of the tangent line if I had the correct $f^{\prime }(x)$

use implicit differentiation to find $\frac{dy}{dx}$ in terms of $x$ and $y$

1.) $x^3-<!--\; -\; -->xy+y^2=4$

2.) $y=\sin <!--\; \; -->(xy)$
find $\frac{d^{2}y}{d{x}^2}$

3.) $x^2{y}^2-<!--\; -\; -->2x=3$

$f(x$) = $x^5-<!--\; -\; -->15x^4-<!--\; -\; -->10x^2+20$
Prove that the equation $f(x)=0$ has a real solution with $x\ge <!--\; \ge \; -->1$.
Idea: Started off by coming up with a boundary $[1,x]$ for some real number $x$. $f(1)=-<!--\; -\; -->4$ so is less than zero, then attempted to find an x that gave me a positive result, very quickly came to the conclusion that any number above 1 when inserted into a function gave me a negative number. Pretty sure I could be missing something really simple. Any help or hints would be appreciated.

I've been stuck on a certain implicit differentiation problem that I've tried several times now.
$\frac{x^2}{x+y}=y^2+6$
I know to take the derivatives of both sides and got:
$\frac{(x+y)2x-<!--\; -\; -->(1-<!--\; -\; -->\frac{dy}{dx})x^2}{(x+y{)}^2}=2y\frac{dy}{dx}$
I blackuced that to get:
$2x^2+2xy-<!--\; -\; -->x^2-<!--\; -\; -->x^2\ast <!--\; \ast \; -->dy/dx=(2y\ast <!--\; \ast \; -->dy/dx)(x+y{)}^2$
I then divided both sides by (2y*dy/dx) and multiplied each side by the reciprocals of the first three terms of the left side. Then I factoblack dy/dx out of the left side and multiplied by the reciprocal of what was left to get dy/dx by itself. I ended up with:
$dy/dx=(2y(x+y{)}^2)/(4x^{7}y)$
but this answer was wrong. I only have one more attempt on my online homework and I can't figure out where I went wrong. Please help!

Find $\frac{dy}{dx}$ for $x^2=\frac{x-<!--\; -\; -->y}{x+y}$
I have solved this in two ways.
First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:
$\frac{1-<!--\; -\; -->3x^2-<!--\; -\; -->2xy}{x^2+1}$
So far so good. When I calculated the implicit derivative of the original expression using the quotient rule though, I got a different solution, i.e.:
$-<!--\; -\; -->\frac{x(y+x{)}^2-<!--\; -\; -->y}{x}$
Can anyone explain to me why I get different solutions ?

Given $y=y(x)$ what is
$\frac{d}{dx}(e^x(x^2+y^2))$
I think its the $\frac{d}{dx}$ confusing me, I don't what effect it has compablack to $\frac{dy}{dx}$. Any help will be greatly appreciated.

I am having this equation:
$\frac{1}{x-<!--\; -\; -->a_1}+\frac{1}{x-<!--\; -\; -->a_2}+\cdots <!--\; \cdots \; -->+\frac{1}{x-<!--\; -\; -->a_n}=0$
where $a_1<a_2<\cdots <!--\; \cdots \; --><a_n$ are real numbers.
Now I want to prove with the intermediate value theorem that this equation has $n-<!--\; -\; -->1$ solutions in the real numbers.
My thoughts:
With $a_1<a_2<\cdots <!--\; \cdots \; --><a_n$, you can see that every summand gets smaller than the summand before.
My other thought was that about the $n$-summands, with the intermediate value theorem you know that every zero (point) is in the interval and is located between the $n$-summands. So there are $n-<!--\; -\; -->1$ solutions for this equation!
Questions:
How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)