\(\displaystyle \begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }2e^{(\frac{x}{3})} - 6x + cos(\pi x) - sin(\pi x) + 2\\&\text{We find the slope function: }\frac{(2e^{(\frac{x}{3})})}{3}- \pi cos(\pi x) - \pi sin(\pi x) - 6\\&\text{Plugging in our value of x we then get:}\\&f'(10)=\frac{(2e^{(\frac{(10)}{3})})}{3}- \pi cos(\pi (10)) - \pi sin(\pi (10)) - 6\\&f'(10)=9.55\end{align*}\)

Find the slope of the line tangent to \(\displaystyle v(t)\) when \(\displaystyle t=3\)

\(\displaystyle v(t)=13t^2-12t+e^t-6\)

Now, let's first look at this conceptually. We need to find the slope of a tangent line at a particular value for t. This sounds like a job for a derivative. 

We will first find v'(t), then we will plug in three to find v'(3), which will yield our answer.

First, recall:

1) \(\displaystyle \frac{d}{dx}e^x=e^x\)

2)\(\displaystyle \frac{d}{dx}x^n=(n)x^{n-1}\)

These two rules are all that we need to solve this problem:

\(\displaystyle v(t)=13t^2-12t+e^t-6\)

\(\displaystyle v'(t)=(2)13t-12+e^t=26t-12+e^t\)

So, we have:

\(\displaystyle v'(t)=26t-12+e^t\)

Let's plug in 3...

\(\displaystyle v'(3)=26(3)-12+e^3=86.09\)

Find the slope of the line tangent to the curve of d(t) when t is equal to 0.

\(\displaystyle d(t)=3t^2-6sin(t)+12cos(t)-11t\)

So, we are asked to find the slope of a tangent line at a particular point. To do so, we need to find the derivative of our function, and then evaluate it at the given value of t.

Let's begin by finding d'(t).

To do so, we need to use the power rule, as well as the rules for sine and cosine.

1) Power Rule

\(\displaystyle \frac{d}{dx}x^n=(n)x^{n-1}\) 

2) Sine Rule

 \(\displaystyle \frac{d}{dx}(sin(x))=cos(x)\)

3) Cosine Rule

\(\displaystyle \frac{d}{dx}(cos(x))=-sin(x)\)

Now, we can use all of these to find the derivative of d(t)

\(\displaystyle d(t)=3t^2-6sin(t)+12cos(t)-11t\)

Becomes:

\(\displaystyle d'(t)=6t-6cos(t)-12sin(t)-11\)

Now, we are almost there, but we need to evaluate this derivative when t=0.

\(\displaystyle d'(0)=6(0)-6cos(0)-12sin(0)-11\)

This simplifies quite nicely to:

\(\displaystyle d'(0)=0-6+0-11=-17\)

So, the slope of our tangent line when t=0 is negative 17.

The function is not differentiable at \(\displaystyle x=1\). So the derivative at \(\displaystyle x=1\) does not exist.

The slope of the line tangent to a function is nothing more than the first derivative, which is for this function equal to

\(\displaystyle g'(x)=2x\cos(x)-x^2\sin(x)\)

found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Evaluated at the given point, we find the slope of the tangent line equal to

\(\displaystyle g'(\pi)=2\pi(\cos(\pi))-\pi^2(\sin(\pi))=-2\pi\)

\(\displaystyle \begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=2\text{, the function appears to be increasing.}\\&\text{The derivative is positive.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=4\text{, the function appears to be decreasing.}\\&\text{The derivative is negative.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=3\text{, the function appears to be increasing.}\\&\text{The derivative is positive.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=-0.3\text{, the function appears to be flat.}\\&\text{The derivative is zero}\end{align*}\)

\(\displaystyle \begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=0.2\text{, the function appears to be flat.}\\&\text{The derivative is zero}\end{align*}\)