Calculus 2 Equations: Expert Solutions and More

You've got a job at a company and you will be traveling to a conference with accommodation. You leave home on monday at 07:15 and arrive at 11:07. The next day you drive exactly the same route back. There is less traffic so you start at 8:32 and arrive at 11:01. Show that there is a point on that route you were at, at the exact same time the two days.

Okay, so that is my question. Does anyone know how to go about this one? I think I need to use the intermediate value theorem, but I'm not sure how to! Thanks in advance for tips/solutions.

a) $e^x=2-<!--\; -\; -->x^2$
b) $\sin <!--\; \; -->(x)=x^2-<!--\; -\; -->1$
The intermediate value theorem says if $F$ is a continues function in a closed interval $[A,B]$, and you choose a value $K$ between $f(A)$ and $f(B)$ where $f(A)\ne f(B)$, Then, there is a value between $A$ and $B$, $C$, such that $f(C)=K$.

I tried to solve the first one by trying to do what the intermediate value theorem says. Then I went on by choosing 2 values A and B. But the first equation has a weird behavior and so does the second one.

Their graphs have 2 long vertical lines with a medium space between them, and Choosing values to plug in doesn't return exact numbers, I'm confused, if anyone can give more clarifications about these problems it would be of great help for me.

Question:
Assume that y is a function of x. Find $\frac{dy}{dx}$ for $x^3+{\cos }^3<!--\; \; -->(y)=5x$

My first move, using the chain rule, is:
$3x^2+3{\cos }^2<!--\; \; -->(y)(-<!--\; -\; -->\sin <!--\; \; -->(y))=5$
However the text says this should instead be:
$3x^2+3{\cos }^2<!--\; \; -->(y)(-<!--\; -\; -->\sin <!--\; \; -->(y))\frac{dy}{dx}=5$
I don't understand where or how the $\frac{dy}{dx}$ is being introduced.
To my mind, ${\cos }^3<!--\; \; -->(y)$=$(\cos <!--\; \; -->(y){)}^3$ and so could be interpreted as:
$f(x)=x^3$
$g(x)=\cos <!--\; \; -->(y)$
$F(x)=f(g(x))={\cos }^3<!--\; \; -->(y)$
Then, using the Chain Rule:
$F^{\prime }(x)=f^{\prime }(g(x))\cdot <!--\; \cdot \; -->g^{\prime }(x)=3{\cos }^2<!--\; \; -->(y)(-<!--\; -\; -->\sin <!--\; \; -->(y))$
But apparently that's wrong.

I have some problem with implicit differentiation.
$y\cos <!--\; \; -->x=4x^2+3y^2$
I am now stuck at the following equality and I don't know how to proceed. Can someone help me?
$y(-\sin <!--\; \; -->x)+(\cos <!--\; \; -->x)y^{\prime }=8x+6y{y}^{\prime }$

Given $f:[0,1]\to <!--\; \to \; -->{\mathbb{R}}^2$ such that $f(0)=(1,0),f(1)=(-<!--\; -\; -->1,0)$ and $f$ is continuous with respect to the standard topologies. How can I prove that there exists $x\in <!--\; \in \; -->[0,1]$ such that $f(x)=(0,y)$, for some $y\in <!--\; \in \; -->\mathbb{R}$?
My attempt to a solution was to consider the restriction of $f$ to the set
$A=\{x:f(x)=(y,0)\text{for some}y\in <!--\; \in \; -->\mathbb{R}\}\subset <!--\; \subset \; -->[0,1].$
Since $f$ is continuous then its restriction is continuous.
Since the restriction of $f$ to $A$ is "basically" a map from $A$ to $\mathbb{R}$, it makes sense that I would just need to apply IVT to finish the proof. I'm having issue in formalizing the reasoning that the restriction is "basically" a map from $A$ to $\mathbb{R}$.

Rates of change
I’m having some trouble with part c) of the following questions,
a) What is the rate of change of the area A of a square with respect to its side x?
b) What is the rate of change of the area A of a circle with respect to its radius r?
c) Explain why one answer is the perimeter of the figure but the other answer is not.
So, knowing that if we have a square with side length x, then the area of the square as a function of its side is $A(x)=x^2$. The perimeter as a function of the side is $P(x)=4x$. And the rate of change of the area wrt its side is $\frac{dA}{dx}=2x$. With a circle, the area as a function of the radius is $A(r)=\mathrm{\pi <!--\; \pi \; -->}(r^2)$. And the rate of change of the area wrt its radius is $\frac{dA}{dr}=2\mathrm{\pi <!--\; \pi \; -->}(r)$. The circumference as a function of the radius is also $C(r)=2\mathrm{\pi <!--\; \pi \; -->}(r)$. Therefore it’s the circle that’s the figure with the rate of change of the area wrt its radius equal to its perimeter, and what I saw was that the square had a rate of change of area wrt its side equal to half the perimeter of the square, $\frac{dA}{dx}=2x=\frac{4x}{2}$
I inscribed a circle in a square with radius equal to half the square’s side length and went through the same work and then arrived at this, $A(r)=\mathrm{\pi <!--\; \pi \; -->}(\frac{x}{2}{)}^2=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{x}^2$ and $C(\frac{x}{2})=2\mathrm{\pi <!--\; \pi \; -->}(\frac{x}{2}$ and that $\frac{dA}{dr}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}x$
Somehow in this example, I don’t think it’s correct because the same fact about the rate of change of area wrt radius being equal to perimeter doesn’t hold. I appreciate any help in explaining this, thank you.

Let's consider the continuous function
$f:\mathbb{R}\times <!--\; \times \; -->[a,b]\to <!--\; \to \; -->\mathbb{R}$
Such that $f(x,a)>0$ for all $x$ and there exists $x_b$ with $f(x_b,b)\le <!--\; \le \; -->0$.
Then there exists $c\in <!--\; \in \; -->[a,b]$ such that $f(x,c)\ge <!--\; \ge \; -->0$ for all $x$ and a real value $x_c$c with $f(x_c,c)=0$.
Intuitively this seems true, but I woludn't know how to prove it.

IVP: A function $f$ has the intermediate value property on an interval $[a,b]$ if for all $x<y$ in $[a,b]$ and all $L$ between $f(x)$ and $f(y)$, it is always possible to find a point $c\in <!--\; \in \; -->(x,y)$ where $f(c)=L$

And

IVT: If $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous, and if $L$ is a real number satisfying $f(a)<L<f(b)$ or $f(a)>L>f(b)$ then there exists a point $c\in <!--\; \in \; -->(a,b)$ where $f(c)=L$Both definitions seem very similar to me, what purpose do they serve individually?

I am given the equation: $x^a{y}^b=6$
Using implicit differentiation I find that the derivative of the equation with respect to y gives: $\frac{d}{dx}(y)=-<!--\; -\; -->\frac{ay}{bx}$. However when I attempt to differentiate the "regular" way I don`t seem to reach the same answer. I would greatly appreciate someone walking me through the problem.

So for $f(z)=\frac{1}{z+1}-<!--\; -\; -->\frac{1}{z+4}$, I am supposed to find the Maclaurin Series and give the region which it converges. Also, I am supposed to find $f^{(17)}(0)$ without computing the derivatives.
I know how to find the Maclaurin series, I am just struggling with finding the region it converges and how to compute the $17$th derivative.

Find $dy/dx$ through implicit differentiation:
$\sin <!--\; \; -->(x)=x(1+\tan <!--\; \; -->(y))$
My solution
$\begin{array}{}\text{(1)}& \frac{d}{dx}[\sin <!--\; \; -->(x)]& =\frac{d}{dx}[x(1+\tan <!--\; \; -->(y))]\text{(2)}& \cos <!--\; \; -->(x)& =(1)(1+\tan <!--\; \; -->(y))+x(1+\tan <!--\; \; -->(y){)}^{-<!--\; -\; -->1}({\sec }^2<!--\; \; -->(y))\frac{dy}{dx}\text{(3)}& \cos <!--\; \; -->(x)& =(1+\tan <!--\; \; -->(y))+\frac{x({\sec }^2<!--\; \; -->(y))}{1+\tan <!--\; \; -->(y)}\frac{dy}{dx}\text{(4)}& \cos <!--\; \; -->(x)-<!--\; -\; -->(1+\tan <!--\; \; -->(y))& =\frac{x({\sec }^2<!--\; \; -->(y))}{1+\tan <!--\; \; -->(y)}\frac{dy}{dx}\text{(5)}& \frac{dy}{dx}& =\frac{(\cos <!--\; \; -->(x)-<!--\; -\; -->1-<!--\; -\; -->\tan <!--\; \; -->(y))(1+\tan <!--\; \; -->(y))}{x({\sec }^2<!--\; \; -->(y))}\end{array}$
Solution from manual I'm using
$\begin{array}{}\text{(6)}& \sin <!--\; \; -->(x)& =x(1+\tan <!--\; \; -->(y))\text{(7)}& \cos <!--\; \; -->(x)& =x(se{c}^2(y))y^{\prime }+(1+\tan <!--\; \; -->(y))(1)\text{(8)}& y^{\prime }& =\frac{\cos <!--\; \; -->(x)-<!--\; -\; -->\tan <!--\; \; -->(y)-<!--\; -\; -->1}{x{\sec }^2<!--\; \; -->(y)}\end{array}$

How do you differentiate ${\displaystyle e^{xy}+x^2-y^2=0}$

I am trying to implicitly differentiate the following function:
$\mathrm{\lambda <!--\; \lambda \; -->}=\exp <!--\; \; -->\left[{(\mathrm{\alpha <!--\; \alpha \; -->}+\frac{s}{\mathrm{\lambda <!--\; \lambda \; -->}-<!--\; -\; -->s})}^{-<!--\; -\; -->1}\right]$
Can someone help me with this?

I'm trying to compute the (implicit) derivative $\frac{dy}{dx}$ from the function
$e^{2y}=x^3$
If I use $\ln$ on both sides I can isolate $y$ and find the derivative:
$\ln <!--\; \; -->(e^{2y})=\ln <!--\; \; -->(x^3)$
$2y=3\ln <!--\; \; -->(x)$
$y=\frac{3}{2}\ln <!--\; \; -->(x)$
$y^{\prime }=\frac{3}{2x}$
But if I use implicit differentiation I get:
$\frac{d}{dx}(e^{2y})=\frac{d}{dx}{x}^3$
$\frac{d}{dy}(e^{2y})\frac{d}{dx}(y)=3x^2$
$2e^{2y}\cdot <!--\; \cdot \; -->y^{\prime }=3x^2$
$y^{\prime }=\frac{3x^2}{2e^{2y}}$
I know both methods should give the same result. What am I missing?

As we know, for implicit differentiation we have $\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=-<!--\; -\; -->\frac{F_x}{F_y}$
but if I use a trick that $\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}F}{\mathrm{\partial <!--\; \partial \; -->}x}}{\frac{\mathrm{\partial <!--\; \partial \; -->}F}{\mathrm{\partial <!--\; \partial \; -->}y}}=\frac{\mathrm{\partial <!--\; \partial \; -->}F}{\mathrm{\partial <!--\; \partial \; -->}x}\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}F}=\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}$ as $\mathrm{\partial <!--\; \partial \; -->}F$ cancels each other out, so it is like $\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{F_x}{F_y}$ without the "minus", where did I go wrong?

What do instantaneous rates of change really represent?The derivative of $f(x)$ is the value of the limit of the average rate of change of y with respect to x as the change in x approaches 0. This is the value, in other words, that the average rate of change approaches but NEVER hits.
This means that it is NOT the infintesimal rate of change of y with respect to $dy/dx$ merely approaches the derivative's value. If the rate of change did actually achieve 0 change in x, you'd get 0/0 which is an indeterminant form.
So if the derivative is the literal rate of change at an exact instant -- a rate of change with an interval of 0, what does that actually tell you? Can a specific moment in time really have a rate of change? Is that rate of change ever even maintained, even at a specific instant?
I know that a point by itself can't have a rate of change, you need a continuum of points around it to determine one (hence a limit). What does an instantaneous rate of change tell you?

Let $f$ and $g$ be continuous functions on $[a,b]$ such that $f(a)\ge <!--\; \ge \; -->g(a)$ and $f(b)\le <!--\; \le \; -->g(b)$. Prove $f(x_0)=g(x_0)$ for at least one $x_0$ in $[a,b]$.
Here's what I have so far:
Let $h$ be a continuous function where $h=f-<!--\; -\; -->g$
Since $h(b)=f(b)-<!--\; -\; -->g(b)\le <!--\; \le \; -->0$, and $h(a)=f(a)-<!--\; -\; -->g(a)\ge <!--\; \ge \; -->0$, then $h(b)\le <!--\; \le \; -->0\le <!--\; \le \; -->h(a)$ is true.
So, by IVT there exists some $y$ such that...
And that's where I need help. I read what the IVT is, but I could use some help explaining how it applies here, and why it finishes the proof. Thank you!

$y=x-<!--\; -\; -->a\sin <!--\; \; -->(bx)-<!--\; -\; -->3$
where $x=x(a,b)$
We have to use implicit differentiation to calculate $\frac{dx}{da},\frac{dx}{db}$.
This is what I get.
$\frac{dy}{da}=\frac{dx}{da}-<!--\; -\; -->\sin <!--\; \; -->(bx)$
for $\frac{dx}{da}$ and
$\frac{dy}{db}=\frac{dx}{db}-<!--\; -\; -->ab\cos <!--\; \; -->(bx)\frac{dx}{db}$
for $\frac{dx}{db}$
Can someone please let me know if I am doing it right. Thanks

Solve this limit.
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{(e^x\sin <!--\; \; -->x-<!--\; -\; -->(x+1)\tan <!--\; \; -->x)}{(x\log <!--\; \; -->\cos <!--\; \; -->(x))}$