We define slope as the first derivative of a given function.

Since we have

$\displaystyle f(x)=3x^{2}+7x-5$, we can use the Power Rule  

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$for all $\displaystyle n\neq0$ to determine that

$\displaystyle f'(x)=(2)3x^{2-1}+(1)7x^{1-1}-(0)5=6x+7$ .

We also have a point $\displaystyle (2,6)$ with a $\displaystyle x$-coordinate $\displaystyle 2$, so the slope

$\displaystyle f'(2)=6(2)+7=12+7=19$.

We define slope as the first derivative of a given function.

Since we have 

$\displaystyle f(x)=5x^{2}-9x+14$, we can use the Power Rule  

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$for all $\displaystyle n\neq0$ to determine that 

$\displaystyle f'(x)=(2)5x^{2-1}-(1)9x^{1-1}+(0)14=10x-9$ .

We also have a point $\displaystyle (-4,2)$ with a $\displaystyle x$-coordinate $\displaystyle -4$, so the slope

$\displaystyle f'(-4)=10(-4)-9=-40-9=-49$.

We define slope as the first derivative of a given function.

Since we have 

$\displaystyle f(x)=2x^{2}+13x-7$, we can use the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$for all $\displaystyle n\neq0$ to determine that 

$\displaystyle f'(x)=(2)2x^{2-1}+(1)13x^{1-1}-(0)7=4x+13$ .

We also have a point $\displaystyle (-1,1)$ with a $\displaystyle x$-coordinate $\displaystyle -1$, so the slope

$\displaystyle f'(-1)=4(-1)+13=-4+13=9$.

In order to find the derivative, we need to find $\displaystyle f'(x)$. We can find this by remembering the product rule and knowing the derivative of natural log.

Product Rule:

$\displaystyle [f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)$.

Derivative of natural log:

$\displaystyle f(x)=\ln(g(x))$

$\displaystyle f'(x)=\frac{g'(x)}{g(x)}$

Now lets apply this to our problem.

$\displaystyle f(x)=x^3\ln(x+1)$

$\displaystyle f'(x)=3x^2\ln(x+1)+\frac{x^3}{x+1}$

$\displaystyle f'(0)=3(0)^2\ln(0+1)+\frac{0^3}{0+1}=0$

The derivative of the function is

$\displaystyle f'(z)=\frac{4z\cos(z)-2z^2\sin(z)}{4}$

which was found using the following rules:

$\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$,

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

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

Finally, plug in the point $\displaystyle z=0$ into the first derivative function:

$\displaystyle f'(0)=\frac{0}{4}=0$

The derivative of the function is

$\displaystyle f'(\theta)=3\sin^2(\theta)\cos(\theta)+ e^{\theta}$

and was found using the following rules:

$\displaystyle \frac{d}{dx}(e^{u})=e^{u}\frac{du}{dx}$,

 $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$, 

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

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

To finish the problem, plug in $\displaystyle \theta=2 \pi$ into the derivative function:

$\displaystyle f'(2 \pi)= e^{2\pi}$.

The derivative of the function is

$\displaystyle f'(y)=2\sec^2(y)\tan(y)+\sec^2(y)$

and was found using the following rules:

$\displaystyle \frac{d}{dx}\tan(x)=\sec^2(x)$, $\displaystyle \frac{d}{dx} \sec(x)=\sec(x)\tan(x)$, $\displaystyle \frac{d}{dx} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}$

Then, plug in the point given into the first derivative function:

$\displaystyle f'(2 \pi) = 2(1)(0)+1=1$

We must find the first derivative of the function first:

$\displaystyle f'(x)=-\sin(x)+\frac{e^{\frac{x}{2}}}{2}+3x^2+2x$

The derivative was found using the following rules:

$\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)$, $\displaystyle \frac{d}{dx}(e^u)=e^u\frac{du}{dx}$, $\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}$

Find the second derivative of the function by taking the derivative of the above function:

$\displaystyle f''(x)=-\cos(x)+\frac{e^{\frac{x}{2}}}{4}+6x+2$

An additional rule was used:

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

Now, plug in x=0 into the above function:

$\displaystyle f''(0)=-1+\frac{1}{4}+2=\frac{5}{4}$

To find the second derivative of the function, we first must find the first derivative of the function:

$\displaystyle f'(x)=2e^{2x}+2\cos(3x)-6x\sin(3x)$

The derivative was found using the following rules:

$\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)$, $\displaystyle \frac{d}{dx}e^u=e^u\frac{du}{dx}$, $\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}$, $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

The second derivative is simply the derivative of the first derivative function, and is equal to:

$\displaystyle f''(x)=4e^{2x}-12\sin(3x)-18x\cos(3x)$

One more rule used in combination with some of the ones above is:

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

To finish the problem, plug in x=0 into the above function to get an answer of $\displaystyle 4$.

We define slope as the first derivative of a given function.

Since we have

$\displaystyle f(x)=9x^{2}-12x+5$, we can use the Power Rule

$\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$ to determine that

$\displaystyle f'(x)=(2)9x^{2-1}-(1)12x+(0)5=18x-12$ .

We also have a point $\displaystyle (4,-4)$ with a $\displaystyle x$-coordinate $\displaystyle 4$, so the slope

$\displaystyle f'(4)=18(4)-12=72-12=60$ .