Note that $\displaystyle f$ is defined for all $\displaystyle x>0.$

Using Logarithm Laws we can write $\displaystyle x^{x}}$ as $\displaystyle e^{xln(x))}$.

Using the Chain Rule, and the Product Rule . We have:

 $\displaystyle (xln(x))'=ln(x)+\frac{x}{x}=ln(x)+1$.

Note that,

 $\displaystyle \frac{\mathrm{d e^{u}} }{\mathrm{d} x}=u'(x)e^{u(x))}$$\displaystyle =(ln(x)+1)e^{xln(x))}$.

Now replace $\displaystyle x^{x}=e^{xln(x)}$ and we get our final answer:

$\displaystyle (ln(x)+1)x^x$

This uses the simple Exponential Rule of derivatives. 

$\displaystyle y=x^a$

Mutiply by the value of the exponent to the function, then subtract 1 from the old exponent to make the new exponent. 

The formula is as follows:

$\displaystyle \frac{dy}{dx}=a\cdot x^{a-1}$.

Using our function,

$\displaystyle y=x^2$ where $\displaystyle a=2$ and $\displaystyle a-1=2-1=1$ we get the following derivative,

$\displaystyle \frac{dy}{dx}=2x^1=2x$.

For this problem we need to use the Chain Rule.

The Chain Rule states to work from the outside in. In this case the outside function is $\displaystyle e^{2x}$ and the inside function is $\displaystyle 2x$. The derivative then becomes the outside function times the derivative of the inside function.

Thus, we use the following formula

$\displaystyle y=e^u$

$\displaystyle y{}'=du\cdot e^{u}$.

In our case $\displaystyle u=2x$, and $\displaystyle du=2$.

Therefore the result is,

$\displaystyle y'=2e^{2x}$.

In order to take the first derivative of the polynomial, all we need to know is how to apply the power rule to a simple term with an exponent:

$\displaystyle \frac{d}{dx}(Ax^n)=nAx^{n-1}$

The formula above tells us that to take the derivative of a term with coefficient $\displaystyle A$ and exponent $\displaystyle n$, we simply multiply the term by $\displaystyle n$ and subtract 1 from $\displaystyle n$ in the exponent. With this in mind, we'll take the derivative of the function in the problem term by term:

$\displaystyle f(x)=4x^4-\frac{2}{3}x^3+\frac{3}{2}x^2-17x+43$

$\displaystyle f'(x)=4(4x^{4-1})-3(\frac{2}{3}x^{3-1})+2(\frac{3}{2}x^{2-1})-1(17x^{1-1})+0(43)$

$\displaystyle f'(x)=16x^3-2x^2+3x-17$

We can see that our function involves a series of terms raised to a power, so we will need to apply the power rule as well as the chain rule to find the derivative of the function. First we apply the power rule to the terms in parentheses as a whole, and then we apply the chain rule by multiplying that entire result by the derivative of the expression in parentheses alone:

$\displaystyle f(x)=4(x^3+\frac{5}{2}x^2)^7$

$\displaystyle f'(x)=28(x^3+\frac{5}{2}x^2)^6(3x^2+5x)$

Since we take the derivative with respect to $\displaystyle y$ we only apply power rule to terms that contain $\displaystyle y$. Since all the terms contain $\displaystyle x$, we treat $\displaystyle x$ as a constant.

The derivative of a constant is always $\displaystyle 0$.

So the derivative of 

$\displaystyle w(x)=x^4+9x+5$

is zero.

The derivative of a constant is $\displaystyle 0$.

Recall the power rule that states to multiply by the exponent in front of the constant and then subtract the exponent by 1.

So lets take the derivative of this in sections:

Derivative of $\displaystyle x^4$:

$\displaystyle 4-1=3$

Result is:

$\displaystyle 4x^3$

Lets take the derivative of the next term:

Derivative of $\displaystyle x^1$:

$\displaystyle 1-1=0$

$\displaystyle 1x^0$

The power of something to $\displaystyle 0$ is $\displaystyle 1$.

Our result is:

$\displaystyle 4x^3+1$

For the function

$\displaystyle f(x)=ax^n$

The first derivative is

$\displaystyle f^{'}(x)=anx^{n-1}$

So for

$\displaystyle f(x)=9x^6$

the first derivative is

$\displaystyle f'(x)=9(6)x^{6-1}=54x^5$

To find the derviative of this equation recall the power rule that states: Multiply the exponent in front of the constant and then subtract one from the exponent.

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

We can work individually with each term:

Derivative of 

$\displaystyle x^2$ is,

$\displaystyle 2x^{2-1}=2x^1=2x$

For the next term:

Derivative of $\displaystyle x^1$:

$\displaystyle 1-1=0$

So answer is:

$\displaystyle 1x^0$

Anything to a power of 0 is 1.

For the next term:

Derivative of $\displaystyle 1$:

Any derivative of a constant is $\displaystyle 0$.

So the first derivative of

$\displaystyle x^2+x+1$

is

$\displaystyle 2x+1$