(a) First we show that $\overline{A^c} \subseteq A^{\circ c}$. If $x \in \overline{A^c}$, then either $x \in A^c$ or $x$ is an accumulation point of $A^c$. If $x \in A^c$, then $x$ is certainly not in the interior of $A$; that is, $x \in A^{\circ c}$. On the other hand, if $x$ is an accumulation point of $A^c$, then every $\epsilon$-neighborhood of $x$ contains points of $A^c$. This means that no $\epsilon$-neighborhood of $x$ lies entirely in $A$. So, in this case too, $x \in A^{\circ c}$.

For the reverse inclusion suppose $x \in A^{\circ c}$. Since $x$ is not in the interior of $A$, no $\epsilon$-neighborhood of $x$ lies entirely in $A$. Thus either $x$ itself fails to be in $A$, in which case $x$ belongs to $A^c$ and therefore to $\overline{A^c}$, or else every $\epsilon$-neighborhood of $x$ contains a point of $A^c$ different from $x$. In this latter case also, $x$ belongs to the closure of $A^c$.

Let $a \in \mathbb{R}$. Given $\epsilon>0$, choose $\delta=\epsilon / 5$. If $|x-a|<\delta$, then $|f(x)-f(a)|=5|x-a|<5 \delta=\epsilon$.

Suppose there exists a nonempty set $U$ which is properly contained in $A$ and which is both open and closed in $A$. Then, clearly, the sets $U$ and $U^c$ (both open in $A$ ) disconnect $A$. Conversely, suppose that $A$ is disconnected by sets $U$ and $V$ (both open in $A$ ). Then the set $U$ is not the null set, is not equal to $A$ (because $V$, its complement with respect to $A$, is nonempty), is open in $A$, and is closed in $A$ (because $V$ is open in $A$ ).

By the Mean Value Theorem, there exists $c \in (0,M)$ such that $$f'(c)=\frac{f(M)-f(0)}{M-0}=\frac{M^3-0^3}{M} = M^2.$$ Since $f(x)=x^3$, we have $f'(x)=3x^2$. So, we have $f'(c)=3c^2=M^2$. Solving for $c$ gives $c=\frac{M}{\sqrt{3}}$.

Therefore, $a=\sqrt{3}$ and $c=\frac{M}{\sqrt{3}}$.

By the Mean Value Theorem, there exists $c \in (0,2)$ such that $$f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{(2^4+2+3)-(0^4+0+3)}{2}=17.$$ Since $f(x)=x^4+x+3$, we have $f'(x)=4x^3+1$. So, we have $f'(c)=4c^3+1=17$. Solving for $c$ gives $c= \sqrt[3]{4}$.

Therefore, $p=\frac{\ln(4)}{\ln(2)}$. Hence, $c=2^p=\boxed{2^{\frac{\ln(4)}{\ln(2)}}}$.

By the Mean Value Theorem, there exists $c \in (4,16)$ such that $$f'(c)=\frac{f(16)-f(4)}{16-4}=\frac{\sqrt{16}-\sqrt{4}}{12}=\frac{1}{3}.$$ Since $f(x)=\sqrt{x}$, we have $f'(x)=\frac{1}{2\sqrt{x}}$. So, we have $f'(c)=\frac{1}{2\sqrt{c}}=\frac{1}{3}$. Solving for $c$ gives $c= \frac{9}{4}$.

Therefore, $c=\boxed{\frac{9}{4}}$.

To find the 117th derivative of $f(x) = 5 \sin x + 3 \cos x$ evaluated at $x = \pi$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= 5\cos x – 3\sin x \ f”(x) &= -5\sin x – 3\cos x \ f”'(x) &= -5\cos x + 3\sin x \ f^{(4)}(x) &= 5\sin x – 3\cos x \ f^{(5)}(x) &= 5\cos x + 3\sin x \ f^{(6)}(x) &= -5\sin x + 3\cos x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 117th derivative of $f(x)$:

\begin{align*} f^{(117)}(x) &= f^{(4\cdot29+1)}(x) \ &= f^{(1)}(x) \ &= 5\cos x – 3\sin x \ \end{align*}

Therefore, evaluating this expression at $x = \pi$, we get:

\begin{align*} f^{(117)}(\pi) &= 5\cos \pi – 3\sin \pi \ &= 5(-1) – 3(0) \ &= -5 \ \end{align*}

So, $f^{(117)}(\pi) = -5$.

To find the 87th derivative of $f(x) = 4 \cos x – 7 \sin x$ evaluated at $x = 0$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= -4\sin x – 7\cos x \ f”(x) &= -4\cos x + 7\sin x \ f”'(x) &= 4\sin x + 7\cos x \ f^{(4)}(x) &= 4\cos x – 7\sin x \ f^{(5)}(x) &= -4\sin x – 7\cos x \ f^{(6)}(x) &= -4\cos x + 7\sin x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 87th derivative of $f(x)$:

\begin{align*} f^{(87)}(x) &= f^{(4\cdot21+3)}(x) \ &= f^{(3)}(x) \ &= 4\sin x + 7\cos x \ \end{align*}

Therefore, evaluating this expression at $x = 0$, we get:

\begin{align*} f^{(87)}(0) &= 4\sin 0 + 7\cos 0 \ &= 0 + 7(1) \ &= 7 \ \end{align*}

So, $f^{(87)}(0) = 7$.

Starting with the given equation $x^2+y^2-1=0$, we take the derivative with respect to $x$ of both sides, using the chain rule for the $y$-term:

$$\frac{d}{dx}(x^2+y^2-1) = \frac{d}{dx}(0)$$

Using the power rule and chain rule, we get:

$$2x + 2y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$, we have:

$$\frac{dy}{dx} = -\frac{x}{y}$$

Taking the derivative of this expression with respect to $x$ using the quotient rule, we get:

\begin{align*} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(-\frac{x}{y}\right)\ &= -\frac{y\frac{d}{dx}(x)-x\frac{d}{dx}(y)}{y^2}\ &= -\frac{y(1\cdot y + x\cdot y’) – x(0\cdot y + 1\cdot y’)}{y^2} \ &= -\frac{y^{\prime}\cdot x}{y^2} – \frac{1}{y} \end{align*}

Simplifying this expression, we get:

$$y^{\prime \prime} = -\frac{1}{y^3}$$

Taking the derivative of $y^{\prime}$ with respect to $x$ using the quotient rule, we have:

\begin{align*} \frac{d^3y}{dx^3} &= \frac{d}{dx}\left(-\frac{y^{\prime}}{y}\right)\ &= -\frac{y\frac{d}{dx}(y’)-y’\frac{d}{dx}(y)}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y – y^{\prime}\cdot y’}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2} \end{align*}

Substituting the values of $y^{\prime\prime}$ and $y^{\prime}$ obtained earlier, we have:

\begin{align*} y^{\prime \prime \prime} &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2}\ &= -\frac{(-1/y^3) \cdot y}{y^2} + \frac{(-x/y^2)\cdot(-x/y)}{y^2}\ &= \frac{x}{y^5} – \frac{1}{y^3}\ &= -\frac{3x}{y^5} \ \ \ \ \ \text{(using } x^2+y^2=1 \text{)} \end{align*}

Therefore, $y^{\prime \prime}=-\frac{1}{y^3}$ and $y^{\prime \prime \prime}=-\frac{3 x}{y^5}$ for the circle $x^2+y^2-1=0$