Calculus 2 Equations: Expert Solutions and More

My equation is $y=x^{y^2}$
I did the ln of both sides, then I tried implicit differentiation. I got
$y^{\prime }=\frac{x^{y^2}{y}^2}{x}.$

Suppose that $(A,\mathrm{\Sigma <!--\; \Sigma \; -->},m)$ is a measure space and $H$ is a linear functional on $L^{\mathrm{\infty <!--\; \infty \; -->}}(A,\mathrm{\Sigma <!--\; \Sigma \; -->},m)$. If
$\mathcal{U}:=\{u:A\to <!--\; \to \; -->\mathbb{R}|u\text{ is measurable, bounded and }{\int <!--\; \int \; -->}_{A}udm=1\}$
and there are functions $u_1,u_2\in <!--\; \in \; -->\mathcal{U}$ such that
$H(u_1)\le <!--\; \le \; -->0\text{and}H(u_2)\ge <!--\; \ge \; -->0,$
then my question is:

Is there a function $u_3\in <!--\; \in \; -->\mathcal{U}$ such that $H(u_3)=0$?

If $4x^2+5x+xy=4$ and $y(4)=-<!--\; -\; -->20$, find $y^{\prime }(4)$ by implicit differentiation.
I implicitly differentiated the equation, but I don't see how I can use $y(4)=-<!--\; -\; -->20$ to my advantage.

Use the properties of integrals to evaluate integral 0 to 1 ${\displaystyle (4+3x^2)dx}$

Solve the given differential equation by using an appropriate substitution.
(x-y)dx+ x dy=0

Verify that the functions ${\displaystyle y_1\text{and}{y}_2}$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions? ${\displaystyle (1-x\cot x)y{}^{″}-x{y}^{\prime }+y=0,0<x<\pi ;y_1\left(x\right)=x,y_2\left(x\right)=\sin x}$

What exactly is steady-state solution?
In solving differential equation, one encounters with steady-state solution. My textbook says that steady-state solution is the limit of solutions of (ordinary) differential equations when ${\displaystyle t\Rightarrow \mathrm{\infty }}$. But the steady-state solution is given as f(t), and this means that the solution is a function of t - so what is this t being in limit?

Find the point(s) on the surface ${\displaystyle z=xy+\frac{1}{x}+\frac{1}{y}}$ which the tangent plane is horizontal.

Rates of change, compounding rates and exponentiationI have a very (apologies if stupidly) simple question about rates of change that has been bugging me for some time. I can't work out whether it relates to my misunderstanding what a rate of change is, to my misapplying the method for calculating a rate of change or something else. I'm hoping somebody on here can help.
For how I define a rate of change, take as an example a population of 1000 items (e.g. bacteria). I observe this population and after an hour I count the size of the population and see that it has increased by 10% (to 1100). I might hypothesise that the population is growing at the rate of 10% per hour, and if, an hour later, I see that it has grown by 10% again (to 1,210) then I might decide to conclude that it is growing at 10% per hour.
So, a rate of change of "proportion x per hour" means "after one hour the population will have changed by proportion x". If, after 1 hour, my population of bacteria was not 1,100, and if not 1,210 after 2 hours, that would mean that the rate of change was not 10% per hour.
First question: Is this a fair definition of a rate of change?
So far so good and it's easy to calculate the population after any given time using a compound interest-type formula.
But whenever I read about continuous change something odd seems to happen. Given that "grows at the rate of 10% per hour" means (i.e. is just another way of saying) "after 1 hour the original population will have increased by 10%", why do textbooks state that continuous change should be measured by the formula:
$P=P_0{e}^{rt}$
And then give the rate of change in a form where this seems to give the wrong answer (i.e. without adjusting it to account for the continuously compounded growth)? I've seen many texts and courses where 10% per day continuous growth is calculated as (for my above example, after 1 day):
$1000\ast <!--\; \ast \; -->e^{1\ast <!--\; \ast \; -->0.1}=1105.17$
This contradicts the definition of a rate of change expressed as "x per unit of time" stated above. If I was observing a population of 1000 bacteria and observed it grow to a population of 1105 after 1 hour I should surely conclude that it was growing at the rate of 10.5% per hour.
I can get the idea of a continuous rate just fine, and it's easy to produce a continuous rate of change that equates to a rate of 10% per day as defined above (that's just ln 1.1). But I struggle to see how a rate of change that means a population grows by 10.5% in an hours means it is growing at 10% per hour. That's like saying if I lend you money at 1% interest per month I'd be charging you 12% per year.
So what's wrong here? Have I got the wrong end of the stick with my definition of a rate of change, would most people interpret a population increase of 10.5% in an hour as a 10% per hour growth rate, or is something else amiss?

How do you find the derivative of ${\displaystyle x{e}^y=y-1}$

Find $\frac{dy}{dx}$ by implicit differentiation. Note that $y$ is a function of $x$.
Given:
$e^{\frac{x}{y}}=x-<!--\; -\; -->y$
My solution:
Take derivative of left-hand side
$\frac{dy}{dx}{e}^{\frac{x}{y}}=(e^{\frac{x}{y}})\left(\frac{y(1)-<!--\; -\; -->(1)y^{\prime }x}{y^2}\right)=(e^{\frac{x}{y}})\left(\frac{y-<!--\; -\; -->y^{\prime }x}{y^2}\right)$
Take derivative of right-hand side
$\frac{dy}{dx}(x)-<!--\; -\; -->\frac{dy}{dx}(y)=1-<!--\; -\; -->1(y^{\prime })=1-<!--\; -\; -->y^{\prime }$
The expression is now
$(e^{\frac{x}{y}})\left(\frac{y-<!--\; -\; -->y^{\prime }x}{y^2}\right)=1-<!--\; -\; -->y^{\prime }$
Multiply both sides by $y^2$
$e^{\frac{x}{y}}y-<!--\; -\; -->y^{\prime }x=y^2-<!--\; -\; -->y^{\prime }{y}^2$
Rearrange variables
$y{e}^{\frac{x}{y}}-<!--\; -\; -->y^2=y^{\prime }x-<!--\; -\; -->y^{\prime }{y}^2$
Factor $y^{\prime }$
$y{e}^{\frac{x}{y}}-<!--\; -\; -->y^2=y^{\prime }(x-<!--\; -\; -->y^2)$
Factor $(x-<!--\; -\; -->y^2)$ out of right-hand side, divide on the left-hand side
$\frac{y{e}^{\frac{x}{y}}-<!--\; -\; -->y^2}{(x-<!--\; -\; -->y^2)}=y^{\prime }$
The correct answer from the textbook is:
$\frac{y^2-<!--\; -\; -->y{e}^{\frac{x}{y}}}{y^2-<!--\; -\; -->x{e}^{\frac{x}{y}}}=y^{\prime }$
Where did I go wrong? Thanks for your help.

Suppose $f:[1,2]\to <!--\; \to \; -->[5,7]$ is continuous. Show that $f(c)=2c+3$ for some $c\in <!--\; \in \; -->[1,2]$.
First note $f(1)=5$ and $f(2)=7$. By the IVT, all values $c\in <!--\; \in \; -->[1,2]$ are hit. I'm just wondering how to put all of these facts together to arrive at f(c)=2c+3 for all $c\in <!--\; \in \; -->[1,2]$.

$x^y=y^x$
find an expression for $dy/dx$ in terms of $y$ and $x$
for the first part is $y{x}^{y-<!--\; -\; -->1}$ and for the second part i got $\ln <!--\; \; -->(y)y^{x}dy/dx$
when solving for dy/dx i got $y{x}^{y-<!--\; -\; -->1})/(\ln <!--\; \; -->(y)y^x)$
Just was wondering if my work is correct?
Sorry for the bad formatting, Im not sure how to make it formatted nice.

Note: I know how to prove this for $f(a)=f(a+100)$, but no clue how to do it for the value that's asked of me.
There's a continuous function $f(x):\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$
With these properties:
1. $f(x)\le 2\text{for}x\ge 200$
2. $f(x)\le 1\text{for}x\le 100$
3. $f(x)=3\text{for}x=150$
I'm supposed to show that there exists an $a$ for which the property $f(a)=(a+200)$ is true. I've tried applying the intermediate value theorem every which way, but at this point I don't know if that's the right way to go.

How do you do implicit differentiation with the quotient rule?

So I got this classic implicit differentiation problem$y=x^x$
I've tried to tackle it by transforming $x^x$ to the equivalent form of $e^{(\ln <!--\; \; -->x)x}$:
$y=x^x$
$y=(e^{\ln <!--\; \; -->x}{)}^x$
$y=e^{(\ln <!--\; \; -->x)x}$
and then proceeded to differentiate it in the following way:
$y^{\prime }=e^{(\ln <!--\; \; -->x)x}\cdot <!--\; \cdot \; -->\ln <!--\; \; -->x\cdot <!--\; \cdot \; -->\frac{1}{x}$

I understand the method used in implicit differentiation, it's just an application of the chain rule. But why can you define $y$ as a function of $x$?
In this equation for example:
$x^2+y^2=1$
$2x+2y{y}^{\prime }=0$
Why isn't it just this?:
$2x+2y=0$
Thank you in advance.

$f(x)=x^2\exp <!--\; \; -->(\sin <!--\; \; -->(x))-<!--\; -\; -->\cos <!--\; \; -->(x)$ on the interval $[0,\mathrm{\pi <!--\; \pi \; -->}/2]$, I have shown the function is continuous and that there is at least one solution on the interval via using IVT, I know I have to find another solution in the interval such that $f(x)<0$, $f(x)>0$ and then $f(x)<0$, just wondering what is the best way to approach, Im sure there must be a more effective way than just to number crunch, could we consider MVT on the interval, many thanks in advance.

What is the derivative of ${\displaystyle y=6xy}$

Use implicit differentiation to calculate $\frac{dy}{dx}$ for $x^2+xy-<!--\; -\; -->6=y^3$