To find the slope at a point of a function, take the derivative of the function.

$\displaystyle y=log_e(x)+5x$

The derivative of $\displaystyle log_a(x)$ is $\displaystyle \frac{1}{x\:ln(a)}$.    

Therefore the derivative becomes,

$\displaystyle \frac{1}{x\:ln(e)}+5=\frac{1}{x}+5$ since $\displaystyle ln(e)=1$.

Now we substitute the given point to find the slope at that point.

$\displaystyle \frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}$

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $\displaystyle f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

 $\displaystyle \\f'(x)=-\frac{1}{2}(x+2)^{-\frac{1}{2}-1}\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+2)\\ f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}\cdot 1\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}$

From here we need to evaluate at the given point $\displaystyle \left(2,\frac{1}{2}\right)$. In this case, only the x value is important, so we evaluate our derivative at x=2 to get$\displaystyle f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$.

To solve this problem, first we need to take the derivative of the function.

 $\displaystyle \\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\\f'(x)=12x^3+4x$

From here we need to evaluate at the given point $\displaystyle (-1,4)$. In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\displaystyle \\f'(-1)=12(-1)^3+4(-1)\\ f'(-1)=12(-1)+4(-1)\\ f'(-1)=-16$.

To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

 $\displaystyle \\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point $\displaystyle (2,1)$. In this case, only the x value is important, so we evaluate our derivative at x=2 to get

$\displaystyle \\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\\ f'(2)=\frac{4+16-2}{36}\\f'(2)=\frac{18}{36}=\frac{1}{2}$.

The derivative of $\displaystyle y=-ln(x)$ is:

$\displaystyle y'=-\frac{1}{x}$

Substitute the point at $\displaystyle x=1$.

$\displaystyle y'=-\frac{1}{x}=-\frac{1}{1}=-1$

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

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

$\displaystyle f'(x)=g'(x)h(x) + g(x)h'(x)$

Following that procedure, we set $\displaystyle ln(x)$ equal to $\displaystyle g(x)$ and $\displaystyle x^2$ equal to $\displaystyle h(x)$.

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

which can be simplified to

$\displaystyle f'(x)=x+2xln(x) = x(1+2ln(x))$.

Now plug in 1 to find the slope at x=1.

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

Remember that $\displaystyle ln(1)=0$.

To solve for the derivative of $\displaystyle 2y^3=3x^2$, use implicit differentiation which means to take the derivative of each term with respect to the variable in that term.

$\displaystyle 6y^2 \cdot \frac{dy}{dx}=6x$

$\displaystyle \frac{dy}{dx}=\frac{6x}{6y^2}$

Substitute the point into the derivative.

$\displaystyle \frac{dy}{dx}=\frac{6x}{6y^2}=\frac{6(3)}{6(2)^2}=\frac{3}{4}$

We must take the derivative $\displaystyle \small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get 

$\displaystyle \small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$, and on the right side we'll get $\displaystyle \small 2x$.

We include the $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\displaystyle \small y$ is a function of $\displaystyle \small x$, so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get 

$\displaystyle \small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\displaystyle \small 3y^2-1/y$ in order to solve for $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$.

Doing this gives you

 $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$.

We want to find the slope at $\displaystyle \small (1,1)$, so we can sub in $\displaystyle \small 1$ for $\displaystyle \small x$ and $\displaystyle \small y$. 

$\displaystyle \small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$.

Here, we must use the product rule.

Assuming $\displaystyle f(x)=sin(x)$ and $\displaystyle g(x)= ln(x)$, our expression becomes

$\displaystyle cos(x)ln(x)+\frac{sin(x)}{x}$.

The question asks us to evaluate this at $\displaystyle x=\frac{\pi}2$.

$\displaystyle cos\left(\frac{\pi}{2}\right)ln\left(\frac{\pi}{2}\right)+\frac{sin(\frac{\pi}{2})}{\frac{\pi}{2}}=\frac{2}{\pi}$

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $\displaystyle f(x)=2x^{2}-4x+7$ and substitute in the $\displaystyle x$-value of the point $\displaystyle (5,10)$. 

Using the Power Rule $\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}$ for all $\displaystyle n\neq0$, $\displaystyle f'(x)=(2)2x^{2-1}-(1)4x^{1-1}+(0)7=4x-4$. Subbing in $\displaystyle x=5$, $\displaystyle f'(5)=4(5)-4=20-4=16$.