Final, Winter 2020

You should recognize the piecewise function $f(x)$ as being made from a parabola $x^2 + 6x + 12$ and a line $ax + b$.

For $x < -2$, the function $f(x)$ follows a parabola. Here's a visual example:

The parabola stops at the green line where $x = -2$. Let’s find the corresponding point by substituting $x = -2$ into the equation:

$$\begin{array}{rl}{\displaystyle f(-2)}& {\displaystyle =(-2{)}^2+6(-2)+12}\\ {\displaystyle }& {\displaystyle =4-12+12}\\ {\displaystyle }& {\displaystyle =4}\end{array}$$
\begin{aligned}
f(-2) &= (-2)^2 + 6(-2) + 12 \\
&= 4 - 12 + 12 \\
&= 4
\end{aligned}


Thus, the point is $(-2, 4)$.

Next, let's consider the line $ax + b$. It could look something like the following:

A function is differentiable when it’s smooth, with no sharp turns or cusps. Here's a key insight:

A 'sharp turn' is called a cusp.

To avoid a cusp at $x = -2$, the slope of the line must match the slope of the parabola at that point. Let’s calculate the slope by taking the derivative of the parabola and evaluating it at $x = -2$:

$$\begin{array}{rl}{\displaystyle f^{\mathrm{\prime }}(x)}& {\displaystyle =2x+6}\\ {\displaystyle f^{\mathrm{\prime }}(-2)}& {\displaystyle =2(-2)+6}\\ {\displaystyle }& {\displaystyle =-4+6}\\ {\displaystyle }& {\displaystyle =2}\end{array}$$
\begin{aligned}
f'(x) &= 2x + 6 \\
f'(-2) &= 2(-2) + 6 \\
&= -4 + 6 \\
&= 2
\end{aligned}




Now, we know that the line must pass through the point $(-2, 4)$ with a slope of 2. Using the point-slope form of a line:

$$(y-y_1)=m(x-x_1),$$
(y - y_1) = m(x - x_1),


we substitute $m = 2$ and the point $(-2, 4)$:

$$\begin{array}{rl}{\displaystyle (y-4)}& {\displaystyle =2(x+2)}\\ {\displaystyle y-4}& {\displaystyle =2x+4}\\ {\displaystyle y}& {\displaystyle =2x+8}\end{array}$$
\begin{aligned}
(y - 4) &= 2(x + 2) \\
y - 4 &= 2x + 4 \\
y &= 2x + 8
\end{aligned}


So, the final equation of the line is:

$$\boxed{{\displaystyle y=2x+8}}$$
\boxed{y = 2x + 8}


Therefore, the values of the constants are $a = 2$ and $b = 8$.

Let's start from left to right.

The first place to look is where $x=-4$. But this is just a derivative of 0, so that's totally fine.

Next we look at $x=-3$. This is a 'cusp' since it's a 'sharp turn' or 'corner'.

Any 'sharp turn' in a function are called a cusp and the derivative does not exist there.

So at $x=-3$ we know the derivative does not exist.

Next we look at $x=-1$. Since the function is not even continuous there, we therefore also know that the function is not differentiable there.

So we add $x=-1$ to our list.

The next point in question is when $x=1$. Since it looks like the slope is completely vertical at $x=1$, that makes the slope essentially infinite. That means the derivative does not exist there!

The derivative does not exist for any $x$ value where the function is totally vertical.

Moving on to the next point in question would be where $x=4$. The function is not continuous there so therefore it is also not differentiable.

This is asking for all the $x$ values where the slope is strictly less than 0.

From the graph we can see the slope is negative starting when $x = -1$ and continues to be negative until $x=2$.

There are two important exceptions when $x=-1$ and when $x=1$. The derivative does not exist at either of those points, so we can't include them in our final answer. Let's explain this:

When $x=-1$ we have no left side limit, so the derivative does not exist.

When $x=1$ the derivative does not exist because the left hand limit and right hand limit both approach infinity.

So for our final answer we have only:
$$\{x\mid -1<x<2\text{ and }x\mathrm{\ne }1\}$$
\{ x \mid -1 < x < 2 \text{ and } x \neq 1 \}

Here we are looking for $x$ values where the second derivative is 0.

NOTE: These points do not necessarily indicate an inflection point! For instance take the parabola $y=x^4$ when $x=0$.

From the graph we can see an inflection point at $x=-4$ together with a tangent line of finite slope (in this case the slope is 0), so the second derivative must be 0 there.

We also definitively have an inflection point when $x=5$ since the graph changes from concave up to concave down together with a tangent line. So for sure we have $f''(x)=0$ here at $x=5$ too.

We can't really tell what's happening when $x=2$, but for this case the answer key will almost certainly not include this point. It's a local min which can indeed sometimes have $f''(x)=0$ (an example was given earlier with $f(x)=x^4$ at $x=0$).

Careful with $x=1$. Since the first derivative doesn't exist there, the second derivative can't exist either. So we have $f''(x) =0$ only at $x = -4$ and $x=5$.

Final solution is:
$$\{x\mid x=-4,x=5\}$$
\{ x \mid x = -4, x=5 \}


or just :
$$\{-4,5\}$$
\{-4, 5 \}

Here we are looking for when the graph $f(x)$ is concave up.

We can see a small concave up segment from when $x$ is between $-4$ and $-3$.

Then it looks like another small segment when $x$ is between 1 and 4.

Then finally there is a small interval where $f(x)$ is concave up right near then end when $x$ is between 4 and 5.

Note that x=4 is not included.

The absolute minimum can clearly be seen when $x=2$.

The derivative of $\cosh(u)$ with respect to $u$ is:

$$\frac{d}{du}\cosh (u)=\sinh (u)$$
\frac{d}{du} \cosh(u) = \sinh(u)




We are differentiating $\cosh(2x)$, so we need to apply the chain rule because we have a function inside another function (i.e., $2x$ inside $\cosh$).

According to the chain rule, if you have a composite function $\cosh(g(x))$, its derivative is:

$$\frac{d}{dx}\cosh (g(x))=\sinh (g(x))\cdot g^{\mathrm{\prime }}(x)$$
\frac{d}{dx} \cosh(g(x)) = \sinh(g(x)) \cdot g'(x)




Here, $g(x) = 2x$, so:

$$\frac{d}{dx}\cosh (2x)=\sinh (2x)\cdot \frac{d}{dx}(2x)$$
\frac{d}{dx} \cosh(2x) = \sinh(2x) \cdot \frac{d}{dx}(2x)




The derivative of $2x$ with respect to $x$ is:

$$\frac{d}{dx}(2x)=2$$
\frac{d}{dx}(2x) = 2




Now, substitute the derivative of $2x$ into the expression:

$$\frac{d}{dx}\cosh (2x)=\sinh (2x)\cdot 2$$
\frac{d}{dx} \cosh(2x) = \sinh(2x) \cdot 2


Thus, the derivative of $\cosh(2x)$ is:

$$\boxed{{\displaystyle \frac{d}{dx}\cosh (2x)=2\sinh (2x)}}$$
\boxed{\frac{d}{dx} \cosh(2x) = 2\sinh(2x)}




$$\boxed{{\displaystyle \frac{d}{dx}\cosh (2x)=2\sinh (2x)}}$$
\boxed{\frac{d}{dx} \cosh(2x) = 2\sinh(2x)}

To find the derivative of:

$$\frac{\arcsin (x)}{\arccos (x)},$$
\frac{\arcsin(x)}{\arccos(x)},


we will apply the quotient rule.

The quotient rule states that for a function of the form $\frac{f(x)}{g(x)}$, the derivative is:



Here, $f(x) = \arcsin(x)$ and $g(x) = \arccos(x)$.



We need the derivatives of both $f(x)$ and $g(x)$:

- The derivative of $f(x) = \arcsin(x)$ is:

$$f^{\mathrm{\prime }}(x)=\frac{1}{\sqrt{1-x^2}}\mathrm{.}$$
f'(x) = \frac{1}{\sqrt{1 - x^2}}.


- The derivative of $g(x) = \arccos(x)$ is:

$$g^{\mathrm{\prime }}(x)=\frac{-1}{\sqrt{1-x^2}}\mathrm{.}$$
g'(x) = \frac{-1}{\sqrt{1 - x^2}}.




Substitute the functions and their derivatives into the quotient rule:







Simplify the numerator:

$$\frac{\arccos (x)}{\sqrt{1-x^2}}+\frac{\arcsin (x)}{\sqrt{1-x^2}}\mathrm{.}$$
\frac{\arccos(x)}{\sqrt{1 - x^2}} + \frac{\arcsin(x)}{\sqrt{1 - x^2}}.


Using the identity $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$, the expression becomes:

$$\frac{\frac{\pi }{2}}{\sqrt{1-x^2}}\mathrm{.}$$
\frac{\frac{\pi}{2}}{\sqrt{1 - x^2}}.



Using the identity $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$ is a key trick that simplifies many trigonometric expressions involving inverse trigonometric functions.





Thus, the derivative of $\frac{\arcsin(x)}{\arccos(x)}$ is:

$$\boxed{{\displaystyle \frac{\pi }{2\sqrt{1-x^2}}}}\mathrm{.}$$
\boxed{\frac{\pi}{2\sqrt{1 - x^2}}}.

The expression is $e^{(x^e)} + x^{(e^x)}$, which contains two terms:

1. The first term is $e^{(x^e)}$. This is an exponential function where the base is $e$ and the exponent is $x^e$.
2. The second term is $x^{(e^x)}$. Here, the base is $x$, and the exponent is $e^x$.

For differentiation, we need to treat each term carefully using rules such as the chain rule and logarithmic differentiation.



The first term is $e^{(x^e)}$. Applying the chain rule:

$$\frac{d}{dx}{e}^{(x^e)}=e^{(x^e)}\cdot \frac{d}{dx}(x^e)\mathrm{.}$$
\frac{d}{dx} e^{(x^e)} = e^{(x^e)} \cdot \frac{d}{dx}(x^e).


Now, differentiate $x^e$. Since $e$ is a constant, the derivative of $x^e$ is:

$$\frac{d}{dx}(x^e)=e\cdot x^{e-1}\mathrm{.}$$
\frac{d}{dx}(x^e) = e \cdot x^{e - 1}.


So, the derivative of the first term becomes:

$$\frac{d}{dx}{e}^{(x^e)}=e^{(x^e)}\cdot e\cdot x^{e-1}\mathrm{.}$$
\frac{d}{dx} e^{(x^e)} = e^{(x^e)} \cdot e \cdot x^{e - 1}.




The second term is $x^{(e^x)}$. To differentiate this, we use logarithmic differentiation. Let:

$$y=x^{(e^x)}\hspace{0.25em} ⟹\hspace{0.25em} \ln y=e^x\ln x\mathrm{.}$$
y = x^{(e^x)} \implies \ln y = e^x \ln x.


Now differentiate both sides implicitly:

$$\frac{1}{y}\cdot \frac{dy}{dx}=e^x\ln x+\frac{e^x}{x}\mathrm{.}$$
\frac{1}{y} \cdot \frac{dy}{dx} = e^x \ln x + \frac{e^x}{x}.


Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx}=x^{(e^x)}\cdot (e^x\ln x+\frac{e^x}{x})\mathrm{.}$$
\frac{dy}{dx} = x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right).




Now, combine the derivatives of both terms:

$$\begin{array}{rl}{\displaystyle }& {\displaystyle \frac{d}{dx}(e^{(x^e)}+x^{(e^x)})}\\ {\displaystyle }& {\displaystyle =e^{(x^e)}\cdot e\cdot x^{e-1}}\\ {\displaystyle }& {\displaystyle +x^{(e^x)}\cdot (e^x\ln x+\frac{e^x}{x})}\end{array}$$
\begin{aligned}
&\frac{d}{dx} \left( e^{(x^e)} + x^{(e^x)} \right) \\
&= e^{(x^e)} \cdot e \cdot x^{e - 1} \\
&\quad + x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right)
\end{aligned}


Thus, the derivative of the expression is:





This question tests your understanding of **differentiation rules for composite and complex functions**. The goal is to assess whether you can correctly apply the chain rule for exponential functions and **logarithmic differentiation** for functions where both the base and exponent are dependent on $x$.

By mastering this problem, you are developing the ability to handle advanced differentiation tasks involving multiple rules, which is essential for success in calculus.

We are tasked with finding the limit:

$$\underset{x\to 0}{\lim }\frac{5^{(1\mathrm{/}x)}}{2^{(1\mathrm{/}{x}^2)}}$$
\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}


This expression involves two exponential terms: $5^{(1/x)}$ and $2^{(1/x^2)}$.



Let’s analyze the behavior of both terms:

1. The term $5^{(1/x)}$: As $x \to 0$, the exponent $1/x$ grows large, making $5^{(1/x)}$ very large.

2. The term $2^{(1/x^2)}$: As $x \to 0$, the exponent $1/x^2$ grows even faster, making $2^{(1/x^2)}$ grow much larger than $5^{(1/x)}$.

This results in an indeterminate form: $\frac{\infty}{\infty}$.



To simplify the expression, we take the natural logarithm. Define:

$$L=\underset{x\to 0}{\lim }\frac{5^{(1\mathrm{/}x)}}{2^{(1\mathrm{/}{x}^2)}}$$
L = \lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}


Taking the natural logarithm of both sides:

$$\ln L=\underset{x\to 0}{\lim }(\ln 5^{(1\mathrm{/}x)}-\ln 2^{(1\mathrm{/}{x}^2)})$$
\ln L = \lim_{x \to 0} \left( \ln 5^{(1/x)} - \ln 2^{(1/x^2)} \right)


Simplifying further:

$$\ln L=\underset{x\to 0}{\lim }(\frac{\ln 5}{x}-\frac{\ln 2}{x^2})$$
\ln L = \lim_{x \to 0} \left( \frac{\ln 5}{x} - \frac{\ln 2}{x^2} \right)




This limit involves terms that grow at different rates as $x \to 0$, making it useful to find a common denominator to better analyze their behavior.



The least common denominator between $x$ and $x^2$ is $x^2$. Rewrite each term with this common denominator:

$$\frac{\ln 5}{x}=\frac{\ln 5\cdot x}{x^2},\frac{\ln 2}{x^2}=\frac{\ln 2}{x^2}\mathrm{.}$$
\frac{\ln 5}{x} = \frac{\ln 5 \cdot x}{x^2}, \quad \frac{\ln 2}{x^2} = \frac{\ln 2}{x^2}.


Now, combine the terms:

$$\frac{\ln 5\cdot x-\ln 2}{x^2}\mathrm{.}$$
\frac{\ln 5 \cdot x - \ln 2}{x^2}.



As $x \to 0$, the term $\ln 5 \cdot x$ vanishes, leaving:

$$\underset{x\to 0}{\lim }\left(\frac{\ln 5\cdot x-\ln 2}{x^2}\right)=\underset{x\to 0}{\lim }\frac{-\ln 2}{x^2}\mathrm{.}$$
\lim_{x \to 0} \left( \frac{\ln 5 \cdot x - \ln 2}{x^2} \right) = \lim_{x \to 0} \frac{-\ln 2}{x^2}.


Since the numerator is a negative constant and the denominator grows large, the limit is:

$$\underset{x\to 0}{\lim }\frac{-\ln 2}{x^2}=-\mathrm{\infty }\mathrm{.}$$
\lim_{x \to 0} \frac{-\ln 2}{x^2} = -\infty.




Since:

$$\ln L\to -\mathrm{\infty }\hspace{0.25em} ⟹\hspace{0.25em} L\to 0$$
\ln L \to -\infty \implies L \to 0




The limit is:

$$\boxed{{\displaystyle \underset{x\to 0}{\lim }\frac{5^{(1\mathrm{/}x)}}{2^{(1\mathrm{/}{x}^2)}}=0}}$$
\boxed{\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}} = 0}




The professor likely asked this question to assess your understanding of indeterminate forms and your ability to apply logarithmic transformations effectively. This problem reinforces the idea that in calculus, growth rates of different functions (like exponential terms) are crucial for evaluating limits.

We are tasked with solving:

$$\underset{x\to 0}{\lim }\frac{\sinh (x)-x}{\sin (x)-x}$$
\lim_{x \to 0} \frac{\sinh(x) - x}{\sin(x) - x}


As $x \to 0$, both $\sinh(x) - x$ and $\sin(x) - x$ approach 0. This results in the indeterminate form $\frac{0}{0}$, making this a case for applying L'Hopital's Rule.

Remember, L'Hopital's Rule applies when the limit results in either $\frac{0}{0}$ or $\frac{\infty}{\infty}$.



We apply L'Hopital's Rule by differentiating the numerator and denominator separately.

1. **Differentiate the numerator** $\sinh(x) - x$:
- The derivative of $\sinh(x)$ is $\cosh(x)$, and the derivative of $x$ is 1. Thus:

$$\frac{d}{dx}(\sinh (x)-x)=\cosh (x)-1$$
\frac{d}{dx} \left( \sinh(x) - x \right) = \cosh(x) - 1


2. **Differentiate the denominator** $\sin(x) - x$:
- The derivative of $\sin(x)$ is $\cos(x)$, and the derivative of $x$ is 1. Thus:

$$\frac{d}{dx}(\sin (x)-x)=\cos (x)-1$$
\frac{d}{dx} \left( \sin(x) - x \right) = \cos(x) - 1




Now, the limit becomes:

$$\underset{x\to 0}{\lim }\frac{\cosh (x)-1}{\cos (x)-1}$$
\lim_{x \to 0} \frac{\cosh(x) - 1}{\cos(x) - 1}


Evaluating both the numerator and the denominator at $x = 0$:

- $\cosh(0) = 1$, so $\cosh(x) - 1 \to 0$.
- $\cos(0) = 1$, so $\cos(x) - 1 \to 0$.

We encounter the indeterminate form $\frac{0}{0}$ again, so we apply L'Hopital's Rule a second time.

Applying L'Hopital's Rule multiple times is perfectly valid when necessary.



1. **Differentiate the numerator** $\cosh(x) - 1$:
- The derivative of $\cosh(x)$ is $\sinh(x)$.

$$\frac{d}{dx}(\cosh (x)-1)=\sinh (x)$$
\frac{d}{dx} \left( \cosh(x) - 1 \right) = \sinh(x)


2. **Differentiate the denominator** $\cos(x) - 1$:
- The derivative of $\cos(x)$ is $-\sin(x)$.

$$\frac{d}{dx}(\cos (x)-1)=-\sin (x)$$
\frac{d}{dx} \left( \cos(x) - 1 \right) = -\sin(x)




Now, the limit becomes:

$$\underset{x\to 0}{\lim }\frac{\sinh (x)}{-\sin (x)}$$
\lim_{x \to 0} \frac{\sinh(x)}{-\sin(x)}


For small $x$, we use the small-angle approximations $\sinh(x) \approx x$ and $\sin(x) \approx x$. Thus:

$$\underset{x\to 0}{\lim }\frac{x}{-x}=-1$$
\lim_{x \to 0} \frac{x}{-x} = -1



For very small values of $x$, $\sin(x)$ and $\sinh(x)$ behave similarly, leading to:
$$\underset{x\to 0}{\lim }\frac{\sinh (x)}{\sin (x)}=1$$
\lim_{x \to 0} \frac{\sinh(x)}{\sin(x)} = 1





Thus, the value of the limit is:

$$\boxed{{\displaystyle -1}}$$
\boxed{-1}




This problem tests your ability to use L'Hopital's Rule effectively and repeatedly. It emphasizes the importance of knowing small-angle approximations for functions like $\sin(x)$ and $\sinh(x)$. Additionally, the exercise aims to develop an intuition for recognizing when two functions behave similarly for small inputs.

We are tasked with solving:

$$\underset{x\to \frac{\pi }{2}}{\lim }\sec (x)-\tan (x)$$
\lim_{x \to \frac{\pi}{2}} \sec(x) - \tan(x)


As $x$ approaches $\frac{\pi}{2}$, both $\sec(x)$ and $\tan(x)$ tend to infinity, creating the indeterminate form $\infty - \infty$. To properly evaluate this limit, we need to manipulate the expression into a form suitable for limit evaluation.

When dealing with indeterminate forms like $\infty - \infty$, it’s helpful to simplify the terms into a single fraction.



We express $\sec(x)$ and $\tan(x)$ using sine and cosine functions:

$$\sec (x)=\frac{1}{\cos (x)},\tan (x)=\frac{\sin (x)}{\cos (x)}\mathrm{.}$$
\sec(x) = \frac{1}{\cos(x)}, \quad \tan(x) = \frac{\sin(x)}{\cos(x)}.


Now, the limit becomes:

$$\underset{x\to \frac{\pi }{2}}{\lim }(\frac{1}{\cos (x)}-\frac{\sin (x)}{\cos (x)})\mathrm{.}$$
\lim_{x \to \frac{\pi}{2}} \left( \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)} \right).


We can factor out $\frac{1}{\cos(x)}$ from both terms:

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{1-\sin (x)}{\cos (x)}\mathrm{.}$$
\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin(x)}{\cos(x)}.




As $x$ approaches $\frac{\pi}{2}$:

- $\sin\left( \frac{\pi}{2} \right) = 1$.
- $\cos\left( \frac{\pi}{2} \right) = 0$.

Thus, the expression approaches the indeterminate form $\frac{0}{0}$, making it suitable for applying L'Hopital's Rule.



We differentiate the numerator and the denominator separately:

1. **Differentiate the numerator** $1 - \sin(x)$:
- The derivative of $1$ is 0, and the derivative of $\sin(x)$ is $\cos(x)$. Thus:

$$\frac{d}{dx}(1-\sin (x))=-\cos (x)\mathrm{.}$$
\frac{d}{dx} \left( 1 - \sin(x) \right) = -\cos(x).


2. **Differentiate the denominator** $\cos(x)$:
- The derivative of $\cos(x)$ is $-\sin(x)$. Thus:

$$\frac{d}{dx}(\cos (x))=-\sin (x)\mathrm{.}$$
\frac{d}{dx} \left( \cos(x) \right) = -\sin(x).




Now, the limit becomes:

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{-\cos (x)}{-\sin (x)}=\underset{x\to \frac{\pi }{2}}{\lim }\frac{\cos (x)}{\sin (x)}\mathrm{.}$$
\lim_{x \to \frac{\pi}{2}} \frac{-\cos(x)}{-\sin(x)} = \lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{\sin(x)}.


As $x \to \frac{\pi}{2}$:

- $\cos\left( \frac{\pi}{2} \right) = 0$.
- $\sin\left( \frac{\pi}{2} \right) = 1$.

Thus, the limit simplifies to:

$$\frac{0}{1}=0.$$
\frac{0}{1} = 0.




The value of the limit is:

$$\boxed{{\displaystyle 0}}\mathrm{.}$$
\boxed{0}.




This problem is designed to test your understanding of **trigonometric identities** and how to handle **indeterminate forms** using L'Hopital's Rule.

To find the equation of the tangent line to a function $f(x)$ at a specific point $x = a$, we use the **point-slope form** of a line:

$$y-f(a)=f^{\mathrm{\prime }}(a)(x-a)$$
y - f(a) = f'(a)(x - a)


Here, $f(x) = \tan(x)$, and we need to find the tangent line at $x = \frac{\pi}{4}$.



First, evaluate the function at $x = \frac{\pi}{4}$:

$$f\left(\frac{\pi }{4}\right)=\tan \left(\frac{\pi }{4}\right)=1.$$
f\left( \frac{\pi}{4} \right) = \tan\left( \frac{\pi}{4} \right) = 1.


Thus, the point on the curve is $\left( \frac{\pi}{4}, 1 \right)$.



The derivative of $f(x) = \tan(x)$ is:

$$f^{\mathrm{\prime }}(x)={\sec }^2(x)\mathrm{.}$$
f'(x) = \sec^2(x).


Now, evaluate the derivative at $x = \frac{\pi}{4}$:


$$f^{\mathrm{\prime }}\left(\frac{\pi }{4}\right)={\sec }^2\left(\frac{\pi }{4}\right)$$
f'\left( \frac{\pi}{4} \right) = \sec^2\left( \frac{\pi}{4} \right)

$$={\left(\frac{1}{\cos \left(\frac{\pi }{4}\right)}\right)}^2$$
= \left( \frac{1}{\cos\left( \frac{\pi}{4} \right)} \right)^2

$$={\left(\frac{1}{\frac{\sqrt{2}}{2}}\right)}^2$$
= \left( \frac{1}{\frac{\sqrt{2}}{2}} \right)^2

$$=2.$$
= 2.



Thus, the slope of the tangent line at $x = \frac{\pi}{4}$ is $2$.

Remember, the slope of the tangent line is the derivative evaluated at the given point.



We have the slope $m = 2$ and the point $\left( \frac{\pi}{4}, 1 \right)$. Using the point-slope form:

$$y-1=2(x-\frac{\pi }{4})\mathrm{.}$$
y - 1 = 2\left( x - \frac{\pi}{4} \right).


Simplify the equation:

$$y-1=2x-\frac{\pi }{2}$$
y - 1 = 2x - \frac{\pi}{2}


$$y=2x-\frac{\pi }{2}+1.$$
y = 2x - \frac{\pi}{2} + 1.


Thus, the equation of the tangent line is:

$$y=2x-\frac{\pi }{2}+1.$$
y = 2x - \frac{\pi}{2} + 1.




The equation of the line tangent to $f(x) = \tan(x)$ at $x = \frac{\pi}{4}$ is:

$$\boxed{{\displaystyle y=2x-\frac{\pi }{2}+1}}\mathrm{.}$$
\boxed{y = 2x - \frac{\pi}{2} + 1}.




This problem helps you practice finding the **equation of a tangent line** using the **point-slope formula**.

From part (a), the equation of the tangent line to $f(x) = \tan(x)$ at $x = \frac{\pi}{4}$ was:

$$y=2x-\frac{\pi }{2}+1.$$
y = 2x - \frac{\pi}{2} + 1.


This equation gives us a linear approximation for $\tan(x)$ around $x = \frac{\pi}{4}$.



Linear approximations are most accurate near the point where the tangent line is calculated. Here, we use the tangent line to estimate $\tan\left( \frac{\pi}{5} \right)$.

1. **Substitute $x = \frac{\pi}{5}$ into the tangent line equation:**

$$y=2\left(\frac{\pi }{5}\right)-\frac{\pi }{2}+1.$$
y = 2\left( \frac{\pi}{5} \right) - \frac{\pi}{2} + 1.


2. **Simplify the expression:**

We start by rewriting the fractions with a common denominator. The least common denominator of $\frac{2\pi}{5}$ and $\frac{\pi}{2}$ is 10:

$$y=\frac{4\pi }{10}-\frac{5\pi }{10}+1.$$
y = \frac{4\pi}{10} - \frac{5\pi}{10} + 1.


Now simplify:

$$y=-\frac{\pi }{10}+1.$$
y = -\frac{\pi}{10} + 1.


Thus, the linear approximation for $\tan\left( \frac{\pi}{5} \right)$ is:

$$y=1-\frac{\pi }{10}\mathrm{.}$$
y = 1 - \frac{\pi}{10}.




The linear approximation of $\tan\left( \frac{\pi}{5} \right)$ based on the tangent line at $x = \frac{\pi}{4}$ is:

$$\boxed{{\displaystyle 1-\frac{\pi }{10}}}\mathrm{.}$$
\boxed{1 - \frac{\pi}{10}}.


Linear approximations work best for values close to the point of tangency. Here, the approximation is most accurate for $x$ near $\frac{\pi}{4}$.

The volume $V$ of a right circular cylinder with height $h$ and radius $r$ is given by the formula:

$$V=\pi r^{2}h$$
V = \pi r^2 h


We are asked to find how fast the volume is changing with respect to time, i.e., $\frac{dV}{dt}$, at a specific moment when:
- $h = 6 \, \text{cm}$ and increasing at a rate of $\frac{dh}{dt} = 2 \, \text{cm/s}$,
- $r = 10 \, \text{cm}$ and decreasing at a rate of $\frac{dr}{dt} = -1 \, \text{cm/s}$.


To find how fast the volume is changing, we differentiate $V = \pi r^2 h$ with respect to time $t$ using the product rule. The radius $r$ and height $h$ are both functions of time, so we apply the chain rule:

$$\frac{dV}{dt}=\pi (2r\frac{dr}{dt}h+r^2\frac{dh}{dt})$$
\frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)


This equation accounts for the rates of change of both the radius and the height.


We are given the following information at the instant of interest:
- $h = 6 \, \text{cm}$,
- $\frac{dh}{dt} = 2 \, \text{cm/s}$,
- $r = 10 \, \text{cm}$,
- $\frac{dr}{dt} = -1 \, \text{cm/s}$.

Substitute these values into the differentiated volume equation:

$$\frac{dV}{dt}=\pi (2(10)(-1)(6)+(10{)}^2(2))$$
\frac{dV}{dt} = \pi \left( 2(10)(-1)(6) + (10)^2(2) \right)



Now, simplify each term in the expression:

$$\frac{dV}{dt}=\pi (2\times 10\times -1\times 6+100\times 2)$$
\frac{dV}{dt} = \pi \left( 2 \times 10 \times -1 \times 6 + 100 \times 2 \right)


$$\frac{dV}{dt}=\pi (-120+200)$$
\frac{dV}{dt} = \pi \left( -120 + 200 \right)


$$\frac{dV}{dt}=\pi (80)$$
\frac{dV}{dt} = \pi (80)


Thus, the rate of change of the volume is:

$$\frac{dV}{dt}=80\pi \hspace{0.17em}{\text{cm}}^3\mathrm{/}\text{s}$$
\frac{dV}{dt} = 80\pi \, \text{cm}^3/\text{s}



The volume of the cylinder is increasing at a rate of:

$$\boxed{{\displaystyle 80\pi \hspace{0.17em}{\text{cm}}^3\mathrm{/}\text{s}}}$$
\boxed{80\pi \, \text{cm}^3/\text{s}}


Use the product rule when differentiating expressions with multiple variables that change with time.