Calculus 2 Equations: Expert Solutions and More

Solving Differential Equations w/ polynomials
I'm new to solving differential equations.
How would we go about solving a differential equation like
$y^{\prime }=-<!--\; -\; -->4+5y-<!--\; -\; -->y^2$?

Use implicit differentiation to determine $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}$ in $yz=ln(x+z)$ and $\sin <!--\; \; -->(xyz)=x+2y+3z$.
Here is my answer:
$yz=ln(x+z)$
$y{z}^{\prime }=(1+z^{\prime })\frac{1}{x+z}$
$z^{\prime }=\frac{1}{yx+yz-<!--\; -\; -->1}$
and
$\sin <!--\; \; -->(xyz)=x+2y+3z$
$y(z+x{z}^{\prime })\cos <!--\; \; -->(xyz)=1+3z^{\prime }$
$yz\cos <!--\; \; -->(xyz)+xy{z}^{\prime }\cos <!--\; \; -->(xyz)=1+3z^{\prime }$
$yz\cos <!--\; \; -->(xyz)-<!--\; -\; -->1=z^{\prime }(3-<!--\; -\; -->xy\cos <!--\; \; -->(xyz))$
$z^{\prime }=\frac{yz\cos <!--\; \; -->(xyz)-<!--\; -\; -->1}{3-<!--\; -\; -->xy\cos <!--\; \; -->(xyz)}$
is that right?

$f:[0;1]\to <!--\; \to \; -->\mathbb{R}$ is continuous and
${\int <!--\; \int \; -->}_0^{1}f(x)dx=\frac{1}{2}$
Proof that there exists an $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=c$.
Idea: So i would define a function $g(x)=f(x)-<!--\; -\; -->x$ and then find points so that $g(x)$ would be $g(y)<0$ and $g(z)>0$. And proceed with my proof and use the intermediate value theorem.
However, i am confused how to do that since i don't have the exact function but only the integral.
Can i just define an $f(x)$ for which the integral would work like $f(x)=\frac{1}{2}$?

How do you find dy/dx for ${\displaystyle y+2x-3x{y}^3=4}$

How do you use implicit differentiation to find ${\displaystyle y\prime }$ for ${\displaystyle \sin \left(xy\right)=1}$?

What is the Maclaurin series of
$z^3\sin <!--\; \; -->(z^2)$

Trying to take a different approach in proving if $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous and injective, then the maximum value of $f$ must occur at $a$ or $b$. trying to prove this by using the intermediate value theorem.
Here is what I have so far:
Let $b\in <!--\; \in \; -->\mathbb{R}$ Assume to the contrary that the max value occurs at an interior point of the interval $[a,b]$. Let $x_{\circ <!--\; \circ \; -->}$ equal this interior point.
The first case would occur when $f(a)<f(b)$ then $f(a)<f(b)<f(x_{\circ <!--\; \circ \; -->})$ If we consider $y_{\circ <!--\; \circ \; -->}\in <!--\; \in \; -->(f(b),f(x_{\circ <!--\; \circ \; -->}))$ is there a way to show that $y_{\circ <!--\; \circ \; -->}$ has at least 2 preimages by applying the intermediate value theorem?
Of course the second case would be $f(b)<f(a)$...

I'm taking differential calculus, and I've read that is actually a pretty interesting function that just has one solution. I'm a little confused since most proofs that involve the Intermediate value theorem give a closed interval. But I need to prove that it has a solution in the real numbers. Thanks in advance.

Given is $F(x,y)=y{e}^{3x}-<!--\; -\; -->2x^2=0$
I was asked to calculate $y^{\prime }$ using implicit differentiation.
I know that $y^{\prime }=-<!--\; -\; -->\frac{Fx}{Fy}=-<!--\; -\; -->\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}.$
So I obtained: $y^{\prime }(x)=-<!--\; -\; -->\frac{3y{e}^{3x}-<!--\; -\; -->4x}{e^{3x}}=\frac{-<!--\; -\; -->3y{e}^{3x}+4x}{e^{3x}}.$
But, in the solution manual I found another approach:
$y=f(x)$
i.e. $f(x)e^{3x}-<!--\; -\; -->2x^2=0$, We use the product rule
$\Rightarrow <!--\; \Rightarrow \; -->f^{\prime }(x)e^{3x}+f(x)3e^{3x}-<!--\; -\; -->4x=0$
$f^{\prime }(x)e^{3x}+f(x)3e^{3x}=4x$
$f^{\prime }(x)e^{3x}=4x-<!--\; -\; -->(f(x)3e^{3x})$
$f^{\prime }(x)=\frac{4x-<!--\; -\; -->(3e^{3x}f(x)}{e^{3x}}$
And finally
$f^{\prime }(x)=y^{\prime }(x)=\frac{4x-<!--\; -\; -->3e^{3x}y}{e^{3x}}=\frac{-<!--\; -\; -->3y{e}^{3x}+4x}{e^{3x}}$
So, the result is the same.

Are both approaches valid? Is there any difference between them? Which one would you recommend to use?

Multivariable Instantaneous rate of change clarification
When you are computing the instantaneous rate of change for f(x,y) what do you take the derivative with respect to?
for example, for
f(x,y)=(sin(πx)cos(πy),yexy,x2+y3)
If I was to find the instantaneous rate of change for all 3 of these functions going through (1,2) with the velocity vector (3,-2) would I just take ddx of all of the functions at (3,-2)?

Could someone explain chain rule when it comes to implicit differentiation?
so basically I want to know why when we have something like:
$v(x)=x-<!--\; -\; -->y+1$
If we take the derivative with respect to x, it yields:
$v^{\prime }(x)=1-<!--\; -\; -->\frac{dy}{dx}$
Now I still don't understand why when it comes to implicit differentiation, we need to tag a $y^{\prime }$ or $\frac{dy}{dx}$ after every time we take the derivative of a y term.

I have seen how people implicitly differentiate the equation $x^2+y^2=c$.
$d/dx(x^2)+d/dx(y^2)=d/dx(c)$
treating "$y$" as "$f(x)$" and using the chainrule we get
$2x+2y(y^{\prime })=0$
and solving for $y^{\prime }$
$y^{\prime }=-<!--\; -\; -->2x/2y$
The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "$x$" and "$y$" on the same side of the equation is what´s bothering me and even if I see "$y$" as a function of "$x$" I just end up imagining
$x^2+(-<!--\; -\; -->x^2+c)=c$
which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they be found in a table. $$\int <!--\; \int \; -->{\ln }^2<!--\; \; -->xdx$$

Show that the following equation has exactly one real root:
$3x+2\cos <!--\; \; -->x+5=0$

So I think the way to approach this problem is to use a combination of the intermediate value theorem and the mean value theorem.
But since we are not given a specific interval, I'm not entirely sure what values of $f(a)$ and $f(b)$ to choose. Am I suppose to choose $f(0)$ and $f(1)$ and see that 0 lies in between them. If so, how exactly can we proceed with the mean value theorem after using the intermediate value theorem. Do we have to check for continuity and the number c, or is there some other way?

Found it:
Let $I$ be a real interval. Let $a,b\in <!--\; \in \; -->I$ such that $(a..b)$ is an open interval. Let $f:I\to <!--\; \to \; -->\mathbb{R}$ be a real function which is continuous on $(a..b)$. Let $k\in <!--\; \in \; -->\mathbb{R}$ lie between $f(a)$ and $f(b)$. That is, either $f(a)<k<f(b)$ or $f(b)<k<f(a)$. Then $\mathrm{\exists <!--\; \exists \; -->}c\in <!--\; \in \; -->(a..b)$ such that $f(c)=k$.

Shouldn't this suppose continuity on the closed interval $[a,b]$? Otherwise consider $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->\mathbb{R}$ such that
$f(x)=\{\begin{array}{ll}\arcsin <!--\; \; -->(x)& \text{-1<x<1}\\ 23& \text{if x=1}\\ -<!--\; -\; -->23& \text{if x= -1}\end{array}$
where $a=-<!--\; -\; -->1$ and $b=1$.

Find ${\displaystyle \frac{dy}{dx}}$ by implicit differentiation.
$\tan <!--\; \; -->(x+y)=x$
So far, I got to
${\displaystyle {\sec }^2<!--\; \; -->(x+y)(1+\frac{dy}{dx})=1}$
but then im lost.. can someone please help and explain? I would really appreciate it!!

Derivative Instantaneous rate of change
Question:
The cumulative ticket sales for the 10 days preceding a popular concert is given by
$S(x)=4x^2+50x+50001\le <!--\; \le \; -->x\le <!--\; \le \; -->10.$
Find the instantaneous rate of change in S(x) when x=3.

Using implicit differentiation, find ${\displaystyle \frac{dy}{dx}}$
$x^3{y}^6=(x+y{)}^9$
My Attempt:
$x^3{y}^6=(x+y{)}^9$
Differentiating both sides with respect to $x$, we get:
${\displaystyle \frac{d}{dx}}(x^3{y}^6)={\displaystyle \frac{d}{dx}}((x+y{)}^9)$
$x^3.{\displaystyle \frac{d}{dx}}{y}^6+y^6.{\displaystyle \frac{d}{dx}}{x}^3=9(x+y{)}^8{\displaystyle \frac{d}{dx}}(x+y)$
$6x^3{y}^5.{\displaystyle \frac{dy}{dx}}+3x^2{y}^6=9(x+y{)}^8(1+{\displaystyle \frac{dy}{dx}})$
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{9(x+y{)}^8-<!--\; -\; -->3x^{2}y}{6x^3{y}^5-<!--\; -\; -->9(x+y{)}^8}}$
But the answer is
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$

The function $g:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$, is continuous, $g(1)>0$ and
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}g(x)=0$
It is a fact that for every $y$ between 0 and $g(1)$ the function takes on a value in $(y,g(1))$
How would one show that if:
$\underset{e\to <!--\; \to \; -->0+}{lim}sup\{g(x):0<x<e\}=0$
then $g$ attains a maximum value on 0 to infinity. Does $g$ necessarily have to be bounded below for this to hold? Why?